It is easy to see that the set of all \(3\times3\) real matrices, which I will denote by \(\mathbb{M}_{3\times3}\) is a vector space. Consider \(H\subset\mathbb{M}_{3\times3}\) defined by
\[H=\{A\in\mathbb{M}_{3\times3}\ \text{such that }\det{(A+A^T)}=0\}.\] Is \(H\) a subspace of \(\mathbb{M}_{3\times3}\)? It turns out, the answer is no. \(H\) has a zero element and is closed under scalar multiplication, but it is not closed under vector addition. It is sufficient to find \(A,B\in H\) with \(A+B\notin H\). Easy, right? Just take \[A=\begin{bmatrix} 1 & 2 & 1\\ 2 & 1 & 2\\ 1 & 2 & 1 \end{bmatrix}\] \[B=\begin{bmatrix} \sqrt{2} & \sqrt{3} & 2\sqrt{2}\\ \sqrt{3} & \sqrt{2} & \sqrt{3}\\ 2\sqrt{2} & \sqrt{3} & \sqrt{2} \end{bmatrix},\] and we are done. \(\square\) As satisfying it would be to just leave these matrices without any trace of where they come from, it contradicts the mantra: a slick write-up is not an instructional one. The first thing to observe is that a matrix plus its transpose is always a symmetric matrix. This is easy to show (addition on real numbers is commutative). So the condition \(\det{(A+A^T)}=0\) is really saying that a certain symmetric matrix related to \(A\) is singular. Moreover, if a matrix is already symmetric, then it is equal to its transpose. This means that the sum of itself with its transpose is equal to two times itself. That is, if \(A\) is a symmetric, \(A+A^T=2A\). But determinants are multilinear (linear on each column vector). So \(\det{2A}=2^n\det{A}\), if \(A\) is an \(n\times n\) matrix. This means that if we choose a symmetric \(A\in\mathbb{M}_{3\times3}\) that is singular, then \(A\in H\), since \(\det{(A+A^T)}=\det{2A}=8\det{A}=0\). In other words, if we choose a symmetric matrix from \(H\), instead of worrying about the sum of that matrix with its transpose, we can just focus on the matrix itself (that is, if \(A\) is symmetric, then \(\det{(A+A^T)}=0\) if and only if \(\det{A}=0\)). So let's investigate symmetric \(3\times3\) matrices. They look like this \[S=\begin{bmatrix} a & b & c\\ b & d & e\\ c & e & f \end{bmatrix}.\] Full disclosure: I stumbled upon this SE post which inspired me to consider the special case of a \(3\times3\) symmetric matrix where \(d=f=a\) and \(e=b\): \[S=\begin{bmatrix} a & b & c\\ b & a & b\\ c & b & a \end{bmatrix}.\] We compute the determinant of such a matrix. \[\begin{split} \det{S}&=a(a^2-b^2)-b(ab-bc)+c(b^2-ac)\\ &=a(a^2-b^2)-b^2(a-c)+b^2c-ac^2\\ &=a^3-b^2a+b^2c-ac^2-b^2(a-c)\\ &=a(a+c)(a-c)-b^2(a-c)-b^2(a-c)\\ &=(a-c)(a^2+ac-2b^2). \end{split}\] In essence, there are three ways that this special case symmetric matrix can be singular. Either only the first factor is equal to zero, only the second factor is equal to zero, or both factors are equal to zero. The sum of two symmetric matrices is also a symmetric matrix, so applying our previous reasoning, instead of worrying about the determinant \(\det{(A+B+(A+B)^T)}\), if we select \(A\) and \(B\) to be symmetric, we need to only worry about \(\det{(A+B)}\). Now watch what happens if we let \(A\) and \(B\) be special-case symmetric matrices in \(H\) with \(A\) having only the first factor in its determinant equal to zero and \(B\) having only the second factor in its determinant equal to zero. In other words, write \[A=\begin{bmatrix} a & b & c\\ b & a & b\\ c & b & a \end{bmatrix}\] \[B=\begin{bmatrix} x & y & z\\ y & x & y\\ z & y & x \end{bmatrix}\] with \(a-c=0\neq a^2+ac-2b^2\), and \(x^2-xz-2y^2=0\neq x-z\). In that case, \[A+B=\begin{bmatrix} p & q & r\\ q & p & q\\ r & q & p \end{bmatrix},\] where \(p=a+x\), \(q=b+y\), and \(r=c+z\). Clearly then it follows that \[p-r=a+x-c-z=x-z\neq0,\] \[p^2-pr-2q^2=(a+x)^2-(a+x)(c+z)-2(b+y)^2=a^2+ac-2b^2+2ax-az-cx-4by=\xi.\] So as long as we choose \(a,b,c,x,y,z\) such that \(\xi\) is nonzero, the determinant of \(A+B\) is guaranteed to be nonzero! This is easy to accomplish. Setting \(a=c=1\) and \(b=2\) satisfies our conditions for the matrix \(A\). Setting \(x=\sqrt{2}\), \(y=\sqrt{3}\), and \(z=2\sqrt{2}\) satisfies our conditions for matrix \(B\), and some quick algebra verifies that \(\xi\neq0\). The result follows! Definitely a fun problem. I reckon there is a simpler solution. My apartment-mate found a counterexample against the closure of \(H\) under addition by trying random things on Matlab. I haven't yet looked into why that solution works, and its corresponding generalization. Stick around for updates!
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The AM-GM inequality is a well-known result, and often a pretty nifty tool. Less well-known is the QM-AM inequality. These two inequalities make up the more general mean-inequality chain. The quadratic mean (QM), sometimes called root-mean-square (RMS), is defined by
\[\sqrt{\frac{x_1^2+...+x_n^2}{n}}.\] The QM-AM inequality then states \[\sqrt{\frac{x_1^2+...+x_n^2}{n}}\geq\frac{x_1+...+x_n}{n}.\] There are two nice ways to prove QM-AM. The first, which I will get out of the way, is by Cauchy-Schwarz. That inequality implies that \[n(x_1^2+...+x_n^2)=(1^2+...+1^2)(x_1^2+...+x_n^2)\geq(x_1+...+x_n)^2.\] Rearranging this yields \[\frac{x_1^2+...+x_n^2}{n}\geq\left(\frac{x_1+...+x_n}{n}\right)^2.\] Both sides are clearly nonnegative, so we can take the square root to obtain the result: \[\sqrt{\frac{x_1^2+...+x_n^2}{n}}\geq\frac{x_1+...+x_n}{n}.\] Cauchy-Schwarz also implies that the equality condition is precisely when the \(x_i\) are all equal. This is clean, but it is pretty opaque. Mostly because Cauchy-Schwarz is not very intuitive (beyond the vector dot product geometric interpretation, do you really have a feel for why the numbers themselves must satisfy Cauchy-Schwarz?). Lucky for us, there is another, arguably more natural, way to derive QM-AM. Suppose I gave you a list of four numbers at McDonald's. For each number, you are allowed to select one of the four types of chicken nuggets box that number of times (and you can only select each box type twice). For instance, if the set I gave you was \(\{2,3,5,8\}\). You select a number, for instance 5, and then select a box type (out of 4-piece, 6-piece, 10-piece, and 20-piece) for instance 10-piece, and buy five 10-pieces to get 50 chicken nuggets. How would you maximize the number of chicken nuggets you get? This is pretty obvious. The greedy algorithm is optimal. We pair the highest box numbers with the highest purchase numbers. This gives rise to what is called the rearrangement inequality. If \(a_i\leq a_j\) and \(b_i\leq b_j\) iff \(i\leq j\), then we have that \(\sum_{i=1}^{n}{a_ib_i}\) is the maximal dot product of vectors with components \(a_i\) with vectors with components \(b_i\). In other words, \[\sum_{i=1}^{n}{a_ib_i}\geq\sum_{k=1}^{n}{a_kb_{\sigma(k)}},\] where \(\sigma\) is a permutation on \(\{1,...,n\}\). There is nothing complicated going on here. Just a super intuitive application of the greedy algorithm. Any third-grader can follow that. Now, it follows that \[a_1b_1+...+a_nb_n\geq a_1b_1+...+a_nb_n\] \[a_1b_1+...+a_nb_n\geq a_1b_2+...+a_nb_1\] \[\vdots\] \[a_1b_1+...+a_nb_n\geq a_1b_{n-1}+...+a_nb_{n-2},\] where in each line we cycle through the order of \(b_i\). Now, observe what happens when we add all \(n\) of the inequalities together. We obtain: \[n\sum_{i=1}^{n}{a_ib_i}\geq\left(\sum_{k=1}^{n}{a_k}\right)\left(\sum_{j=1}^{n}{b_j}\right).\] Perhaps you see where this is going now. Suppose that \(a_i=b_i=x_i\). Then \[n(x_1^2+...+x_n^2)\geq(x_1+...+x_n)^2.\] Now, the rest of the proof follows as before with Cauchy-Schwarz. A practical application of QM-AM is that it immediately establishes that the RMS speed is greater than the average speed of molecules in a gas. This avoids the use of the Maxwell-Boltzmann distribution. Deriving RMS speed can be (and is traditionally) done without Maxwell-Boltzmann, but doing it with Maxwell-Boltzmann is fun because it presents an interesting application of Feynman's trick. Perhaps I'll make a main page post about that. 1994 Wisconsin Mathematics Science & Engineering Talent Search PSET 3 Problem 3: Find all positive integer solutions of the equation
\[x^3-y^3=xy+61.\] Solution: My first instinct is that the LHS factors: \[(x-y)(x^2+xy+y^2)=xy+61.\] I see that there is an \(xy\) term on both sides of the equation. Perhaps if I distribute the \(x-y\) on the LHS and then subtract \(xy\). Then, we obtain \[(x-y)x^2+(x-y-1)xy+(x-y)y^2=61.\] This doesn't do me any good, but I notice that I can factor out \(x-y\), I can obtain \[(x-y)\left(x^2+\left(1-\frac{1}{x-y}\right)xy+y^2\right)=61.\] Aha! Now we can equate the two factors on the left to the integer factors of \(61\). Luckily, \(61\) is prime so there are only four cases to test. The only one that ends up working is \(x-y=1\), which yields \[x^2+(x-1)^2=61\Rightarrow 2x^2-2x-60=0\Rightarrow (x-6)(x+5)=0.\] So our two integer solutions are \((6,5)\) and \((-5,-6)\). The only positive integer solution is thus \(\boxed{(6,5)}\). 1994 Wisconsin Mathematics Science & Engineering Talent Search PSET 3 Problem 4: Without using a calculator or computer, determine which of the two numbers \(31^{11}\) or \(17^{14}\) is larger. Solution 1: First, I observed that \(31=2^5-1\) and \(17=2^4+1\). So the ratio between the two given numbers is \[\begin{split} \frac{31^{11}}{17^{14}}&=\frac{(2^5-1)^{11}}{(2^4+1)^{14}}\\ &=\frac{(2^5-1)^{11}}{(2^4+1)^{14}}\cdot\frac{2^{-56}}{2^{-56}}\\ &=\frac{1}{2}\cdot\frac{(1-2^{-5})^{11}}{(1+2^{-4})^{14}} \end{split}\] Since \(\frac{(1-2^{-5})^{11}}{(1+2^{-4})^{14}}<\frac{(1-2^{-5})^{11}}{1}<\frac{1}{1}\), we must have \(17^{14}>31^{11}\). Solution 2: This time, we proceed with the difference instead of the ratio. \[\begin{split} 17^{14}-31^{11}&=(2^4+1)^{14}-(2^5-1)^{11}\\ &=\sum_{k=0}^{14}{\binom{14}{k}2^{56-4k}}-\sum_{j=0}^{11}{(-1)^j\binom{11}{j}2^{55-5j}}\\ &=\sum_{k=0}^{11}{\binom{14}{k}2^{56-4k}+(-1)^{k+1}\binom{11}{k}2^{55-5k}}+\sum_{j=12}^{14}{\binom{14}{j}2^{56-4j}} \end{split}\] The second summation in the last expression above is positive since each term is positive. At this point, I wonder if I could decompose \(\binom{14}{k}\) into a linear combination of terms of the form \(\binom{11}{r}\). I realize that this is actually possible with Pascal's identity: \[\begin{split} \binom{14}{k}&=\binom{13}{k}+\binom{13}{k-1}\\ &=\binom{12}{k}+2\binom{12}{k-1}+\binom{12}{k-1}\\ &=\binom{11}{k}+3\binom{11}{k-1}+3\binom{11}{k-2}+\binom{11}{k-3} \end{split}\] Curiously, the coefficients themselves are binomial coefficients! The proof of that follows pretty naturally. I bet there is a counting argument for it too. Anyway, our first summation can now be written as: \[\begin{split} \sum_{k=0}^{11}{\binom{14}{k}2^{56-4k}+(-1)^{k+1}\binom{11}{k}2^{55-5j}}&=\sum_{k=0}^{11}{\left[\binom{11}{k}+3\binom{11}{k-1}+3\binom{11}{k-2}+\binom{11}{k-3}\right]2^{56-4k}+(-1)^{k+1}\binom{11}{k}2^{55-5k}}\\ &=\sum_{k=0}^{1}{\binom{11}{k}\left(2^{56-4k}+(-1)^{k+1}2^{55-5k}\right)+\left[3\binom{11}{k-1}+3\binom{11}{k-2}+\binom{11}{k-3}\right]2^{56-4k}} \end{split}\] Observe that for due to the definition of generalized binomial coefficients, \(\binom{11}{r}=0\) for negative integers \(r\). Furthermore, we observe that to have \(2^{56-4k}>2^{55-5k}\), we must have \(56-4k>55-5k\) or \(k>-1\). But this is obviously always true for our index \(k\) which starts at \(0\). Hence, every term in the summations is positive. So we must have \(\boxed{17^{14}>31^{11}}\). Here is a cute problem from Sweden. It is simply stated. Solve
\[(\sqrt{2}-1)^x+(\sqrt{2}+1)^x=6.\] Before reading on, I strongly encourage you to give this a shot yourself. This is one of those problems that make you slap your forehead in disgust when you see the solution because it's so dumb simple. Ok. Here's my solution. We multiply through with, say, \((\sqrt{2}+1)^x\). This gives us \[1^x+(\sqrt{2}+1)^{2x}=6(\sqrt{2}+1)^x.\] But \(1^x=x\forall x\in\mathbb{R}\) so \[1+(\sqrt{2}+1)^{2x}=6(\sqrt{2}+1)^x.\] Aha! Now this is quadratic in \((\sqrt{2}+1)^x\). Letting \(y=(\sqrt{2}+1)^x\), we have \[y^2-6y+1=0,\] from which we find the roots to be \[y=3\pm2\sqrt{2}.\] Hence, \[(1+\sqrt{2})^x=3\pm2\sqrt{2}\] and we have our two solutions \[\boxed{x=\log_{1+\sqrt{2}}{(3\pm2\sqrt{2})}=\pm2}\] Haha... Here are some inequalities I solved today/yesterday.
Problem (USAMTS): Find the ordered pair of real numbers \((x,y)\) that satisfies the equation below, and demonstrate that it is unique: \[\frac{36}{\sqrt{x}}+\frac{9}{\sqrt{y}}=42-9\sqrt{x}-\sqrt{y}.\] Solution: The equation rearranges as: \[\frac{36}{\sqrt{x}}+9\sqrt{x}+\frac{9}{\sqrt{y}}+\sqrt{y}=42.\] Next, observe that by AM-GM: \[\frac{36}{\sqrt{x}}+9\sqrt{x}\geq2\sqrt{36\cdot9}=36,\] and: \[\frac{9}{\sqrt{y}}+\sqrt{y}\geq2\sqrt{9}=6.\] Adding these two inequalities yields: \[\frac{36}{\sqrt{x}}+9\sqrt{x}+\frac{9}{\sqrt{y}}+\sqrt{y}\geq42,\] so our equation is actually just the equality condition of AM-GM applied to \(x\) and \(y\) terms separately. Equality in AM-GM holds only when the terms are equal, hence the solution to the equation is unique. We have: \[\frac{36}{\sqrt{x}}=9\sqrt{x}\Rightarrow\boxed{x=\frac{36}{9}},\] and: \[\frac{9}{\sqrt{y}}=\sqrt{y}\Rightarrow\boxed{y=9}.\] Problem: Show that for all positive real numbers \(x\neq1\) and nonnegative integers \(n\), we have \[\frac{x^{2n+1}-1}{x^{n+1}-x^n}\geq2n+1.\] Solution: Observe that the LHS can be rewritten: \[\begin{split} \frac{x^{2n+1}-1}{x^{n+1}-x^n}&=\frac{x^{2n+1}-1}{x^n(x-1)}\\ &=\frac{x^{n+1}-\frac{1}{x^n}}{x-1}\\ &=\frac{\frac{1}{x^n}(1-x^{2n+1})}{1-x}. \end{split}\] This is the sum of a finite geometric series with first term \(\frac{1}{x^n}\), common ratio \(x\), and \(2n+1\) terms. Therefore, our inequality can be written as \[x^{-n}+x^{-n+1}+...+x^n\geq2n+1.\] But this is true by AM-GM on the LHS. Since all of our steps are reversible, we are done. \(\square\) Problem: Show that for all positive integers \(n\) with \(n\geq2\), we have \[\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n-1}>n(2^{1/n}-1).\] Solution: We rearrange the inequality to: \[\frac{\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n-1}}{n}+1>2^{1/n}.\] This is simply: \[\frac{n+\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n-1}}{n}>2^{1/n}.\] Now, we break up the \(n\) in the numerator into \(n\) ones and allocate the ones to every remaining term in the numerator: \[\frac{\left(1+\frac{1}{n}\right)+\left(1+\frac{1}{n+1}\right)+...+\left(1+\frac{1}{2n-1}\right)}{n}>2^{1/n}.\] But now, this becomes: \[\frac{\frac{n+1}{n}+\frac{n+2}{n+1}+...+\frac{2n}{2n-1}}{n}>2^{1/n},\] which is true by AM-GM. Observe that the inequality is strict since obviously none of the terms in the numerator are equal (which is the equality condition for AM-GM). Since all our steps are reversible, we are done. \(\square\) I didn't find (or solve) any interesting Cauchy-Schwarz problems. :(. Tomorrow/today I think I will work on combo. Here is my current progress on the problem I described in my previous post.
First, I observed that \(|a_{n+1}-a_n|\) is maximized when \(a_n\) is equidistant (or as close as possible to being equidistant) to consecutive multiples of \(n+1\). Hence: \[|a_{n+1}-a_n|\leq\left\lfloor \frac{n+1}{2} \right\rfloor\] Next, I considered a general term \(a_n=np\) and \(a_{n+1}=(n+1)q\) (where of course \(p,q\in\mathbb{Z}\). WLOG, I let \(a_{n+1}>a_n\). This was the case of interest anyway. Under these assumptions, I had: \[np+R=(n+1)q\Rightarrow q=\frac{np+R}{n+1}\] I set this to be less than or equal to \(p\). At the time, it just assumed this via heuristics. It wasn't until later until I was able to actually prove it. Lemma: Let \((n+1)q>np\) be the closest multiple of \(n+1\) to \(np\). Then \(p\geq q\) Proof: Suppose that to the contrary, \((n+1)r>np\) was the closest multiple of \(n+1\) to \(np\), where \(r>p\). Then, the difference between the two numbers is: \[nr+r-np=n(r-p)+r\] Observe that \(r-p\geq1\Rightarrow n(r-p)\geq n\) and \(r>1\) (by implicit assumption). Adding these two inequalities yields: \[n(r-p)+r>n+1\] The maximum possible distance between a multiple of \(n\) and the closest larger multiple of \(n+1\) is \(n+1\), and this occurs when the aforementioned multiple of \(n\) is also a multiple of \(n+1\). However, we have shown that the distance between \((n+1)r\) and \(np\) is greater than \(n+1\), contradicting our initial assumption that \((n+1)r\) was the closest multiple of \(n+1\) that was larger than \(np\). Hence, \(\exists q\leq p\) such that \((n+1)q\) is the closest larger multiple of \(n+1\) to \(np\). \(\square\) Ok that's cool. So that means that we can safely set our expression for \(q\) to be lesser than or equal to \(p\): \[\frac{np+R}{n+1}\leq p\] We can rearrange this inequality to obtain: \[p\geq R\] The equality condition is more interesting. When \(p=R\), we must also have \(p=q\), and the sequence becomes arithmetic (\(n=k\)). This can more easily be obtained by noting that \(R=(n+1)q-np\). But when does \(p=R\)? Here is what I did know: \[0\leq R\leq\left\lfloor \frac{n+1}{2} \right\rfloor\] \(0\) wasn't particularly interesting... but what about the upper bound? Letting \(R\) equal its upper bound seemed to match experimental results of my program for \(a_k\)! So I strongly suspect that: \[a_k=k\left\lfloor \frac{k+1}{2} \right\rfloor\] The question is, does this \(a_k\) always occur? How do I go about showing that it does? I have shown that \(q\leq p\), but how do I show that \(q=p\) will eventually occur? This is from Volume 2 (AoPS). Problem 216. Show that for any two positive real numbers \(a\) and \(b\),
\[\frac{a+b}{2}-\sqrt{ab}\geq\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}\] by showing that this inequality is equivalent to \[\frac{(a+b)^2}{2}\geq\sqrt{2ab(a^2+b^2)}\] and then using the AM-GM Inequality. This is a two part problem, so I will split the solution in two parts. I am posting this here because the algebraic manipulation for part 1 is cute, I used the trivial inequality instead of AM-GM for part 2, and overall, this is too insignificant to merit an entire PDF. I must confess, I solved part 2 before part 1. Part 1: At first, this seems intractable because we need to somehow end up with the square root of a product. After about twenty seconds of "what even is this" staring, I realize that the same term is on both sides of the inequality. OK. Now I have: \[a+b-\sqrt{ab}\geq\sqrt{\frac{a^2+b^2}{2}}\] Things became a lot more apparent once I wrote: \[a+b\geq\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\] Now if I square the inequality and divide by 2, the LHS matches the LHS of our destination! Hence, it suffices to show that the resulting RHS is equivalent to the destination RHS. In particular, we have: \[(a+b)^2\geq\frac{a^2+b^2}{2}+2\sqrt{\frac{1}{2}ab(a^2+b^2)}+ab\] Dividing by two: \[\frac{(a+b)^2}{2}\geq\frac{a^2+b^2}{4}+\sqrt{\frac{1}{2}ab(a^2+b^2)}+\frac{ab}{2}\] Now we can work on the RHS: \[\frac{a^2+b^2}{4}+\sqrt{\frac{1}{2}ab(a^2+b^2)}+\frac{ab}{2}=\frac{a^2+b^2+2ab}{4}+\sqrt{\frac{1}{2}ab(a^2+b^2)}\] Which is: \[\frac{(a+b)^2}{4}+\sqrt{\frac{1}{2}ab(a^2+b^2)}\] At this point, I had didn't really have any ideas as to how to turn this into \(\sqrt{2ab(a^2+b^2)}\). It wasn't until I took a look at the LHS of the inequality once again when I realized that I could subtract \(\frac{(a+b)^2}{4}\) from both sides and then multiply by two to obtain the desired: \[\boxed{\frac{(a+b)^2}{2}\geq\sqrt{2ab(a^2+b^2)}}\] Part 2: When I'm solving a problem that is dependent on clever algebraic manipulations, I essentially think about terms that would result from possible operations in my head. I find this to be a great way to narrow down what operations need to be performed to make any progress. In this case, we have: \[\frac{(a+b)^2}{2}\geq\sqrt{2ab(a^2+b^2)}\] The \(a^2+b^2\) and the \(2ab\) in the radical were just too alarming to ignore. Especially considering that we obtain these very terms in the expansion of the LHS. In fact, I was so certain that these terms would be preserved that I went ahead and let \(x=a^2+b^2\) and \(y=ab\). Hence: \[\frac{x}{2}+y\geq\sqrt{2xy}\] Squaring: \[\frac{x^2}{4}+xy+y^2\geq2xy\] Standard manipulations turns this into: \[x^2-4xy+4y^2\geq0\] But this factors! \[(x-2y)^2\geq0\] Which is true by the Trivial Inequality. \(\square\) I'm not sure what the AM-GM solutions was. My issue is that this solution just seems so overwhelmingly natural... those \(a^2+b^2\) and \(ab\) terms were just begging for substitution. Perhaps I just need a better grasp on nontrivial inequalities... EDIT: Never mind, I just realized that the results follows from AM-GM on: \[\frac{x}{2}+y\geq\sqrt{2xy}\] The reason I did not observe this in my original solution is because I actually bothered to make my substitutions after completely expanding everything to eliminate the radical. This eliminated the radical symbol that is apparently so necessary for my brain to realize that AM-GM is a viable option. I am confident that if I had made my substitution earlier, I would have used AM-GM. Man, after such a long hiatus, I am rusty. |
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