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While Muirhead works, another viable option is a generalization of rearrangement by taking four copies of $\{\sqrt{x},\,\sqrt{y},\,\sqrt{z}\}$. (WLOG $x\geq y\geq z\geq 0$) Since $\{\sqrt{x},\,\sqrt{y},\,\sqrt{z}\}$ is monotonic decreasing, we have

Well by doing what I said this is actually what you get :).....:
$x\sqrt {yz}+y\sqrt {zx}+z\sqrt {xy}\le x\frac {y+z}{2}+ y\frac {x+z}{2}+z\frac {y+x}{2}=xy+yz+zx$

@Xuming Liang
–
It's good to finally see some inequality actions going on here, I've never been interested in inequalities until recently :) Here's one I feel kinda proud for solving it:

$a,b,c$ are positive reals, $ab+bc+ca=1$ prove $\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a }\leq\frac{1}{abc}.$

P.S. Good think on AoPS there's a quick way for you to copy and paste latex..

Since anything provable using Muirhead (which was the instakill solution that I suggested) could be proven using AM-GM, the challenge was to prove my suggested inequalities using AM-GM.

By AM-GM, $\dfrac{2x^2+y^2+z^2}{4}=\dfrac{x^2+x^2+y^2+z^2}{4}\ge\sqrt[4]{x^4y^2z^2}.$

Similarly, $\dfrac{x^2+2y^2+z^2}{4}\ge\sqrt[4]{x^2y^4z^2}$ and $\dfrac{x^2+y^2+2z^2}{4}\ge\sqrt[4]{x^2y^2z^4}$. Note that $\sqrt[4]{x^4y^2z^2}=\sqrt{x^2yz}=x\sqrt{yz},$ which can also apply to the other three radicals.

Dividing through, we have this system of inequalities.

Muirhead states that for any two sets of numbers $a_1,a_2,\ldots a_n$ and $b_1,b_2,\ldots b_n$, if $(a_1,a_2,\ldots a_n)\succ(b_1,b_2,\ldots b_n)$, then $\sum_{sym}x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}\ge \sum_{sym}x_1^{b_1}x_2^{b_2}\cdots x_n^{b_n}$

Search up Majorization if you do not know what $\succ$ (which means majorizes) means.

But Muirhead is basically just a generalized AM-GM. Setting the first set of numbers as $1,0,\ldots 0$ and the second set as $\dfrac{1}{n},\dfrac{1}{n},\ldots \dfrac{1}{n}$, we see that $(1,0,\ldots 0)\succ \left(\dfrac{1}{n},\dfrac{1}{n},\ldots \dfrac{1}{n}\right)$ and using Muirhead's gives us the familiar $n$-term AM-GM.

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## Comments

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TopNewestWhile Muirhead works, another viable option is a generalization of rearrangement by taking four copies of $\{\sqrt{x},\,\sqrt{y},\,\sqrt{z}\}$. (WLOG $x\geq y\geq z\geq 0$) Since $\{\sqrt{x},\,\sqrt{y},\,\sqrt{z}\}$ is monotonic decreasing, we have

$\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x} + \sqrt{y}\cdot\sqrt{y}\cdot\sqrt{y}\cdot\sqrt{y} + \sqrt{z}\cdot\sqrt{z}\cdot\sqrt{z}\cdot\sqrt{z}$ $\geq \sqrt{x}\cdot\sqrt{x}\cdot\sqrt{y}\cdot\sqrt{z} + \sqrt{y}\cdot\sqrt{y}\cdot\sqrt{z}\cdot\sqrt{x} + \sqrt{z}\cdot\sqrt{z}\cdot\sqrt{x}\cdot\sqrt{y}$

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We could apply AM-GM on $\sqrt {yz}\le \frac {y+z}{2}$ which would lead to proving the well known $x^2+y^2+z^2\ge xy+yz+xz$

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It is also true that $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}$

Can you prove that?

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Well by doing what I said this is actually what you get :).....: $x\sqrt {yz}+y\sqrt {zx}+z\sqrt {xy}\le x\frac {y+z}{2}+ y\frac {x+z}{2}+z\frac {y+x}{2}=xy+yz+zx$

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However, it is also true that $(x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy})^3\ge 27x^2y^2z^2$

How about that one?

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How about this one?

$2x\sqrt{yz}+2y\sqrt{zx}+2z\sqrt{xy}\ge \sum_{sym}\sqrt[3]{xy^2z^3}$

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$a,b,c$ are positive reals, $ab+bc+ca=1$ prove $\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a }\leq\frac{1}{abc}.$

P.S. Good think on AoPS there's a quick way for you to copy and paste latex..

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@Michael Tang @Trevor B. Try your hand at this one.

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$1,\frac{1}{2},\frac{1}{2}$ majorizes $1,\frac{2}{3},\frac{1}{3}$, use Muirhead to obtain the result.

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@Bogdan Simeonov

Sorry, but the point was to prove it using AM-GMSince anything provable using Muirhead (which was the instakill solution that I suggested) could be proven using AM-GM, the challenge was to prove my suggested inequalities using AM-GM.

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A solution by Muirhead that I discovered while writing the problem is as follows:

Note that $(2,0,0)\succ \left(1,\dfrac{1}{2},\dfrac{1}{2}\right)$.

Thus, by Muirhead, $2(x^2+y^2+z^2)\ge 2(x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy})$ and the result follows.

The challenge I give you is to try to prove it using AM-GM. It shouldn't be too difficult.

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By AM-GM, $\dfrac{2x^2+y^2+z^2}{4}=\dfrac{x^2+x^2+y^2+z^2}{4}\ge\sqrt[4]{x^4y^2z^2}.$

Similarly, $\dfrac{x^2+2y^2+z^2}{4}\ge\sqrt[4]{x^2y^4z^2}$ and $\dfrac{x^2+y^2+2z^2}{4}\ge\sqrt[4]{x^2y^2z^4}$. Note that $\sqrt[4]{x^4y^2z^2}=\sqrt{x^2yz}=x\sqrt{yz},$ which can also apply to the other three radicals.

Dividing through, we have this system of inequalities.

$\begin{aligned} \dfrac{1}{2}x^2+\dfrac{1}{4}y^2+\dfrac{1}{4}z^2&\ge x\sqrt{yz}\\ \dfrac{1}{4}x^2+\dfrac{1}{2}y^2+\dfrac{1}{4}z^2&\ge y\sqrt{zx}\\ \dfrac{1}{4}x^2+\dfrac{1}{4}y^2+\dfrac{1}{2}z^2&\ge z\sqrt{xy} \end{aligned}$

Adding these together yields $x^2+y^2+z^2\ge x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy},$ as desired.

$\mathbb{Q.E.D.}$

@Daniel Liu, Can you please explain your Muirhead solution a little further? I'm not familiar with that.

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Muirhead states that for any two sets of numbers $a_1,a_2,\ldots a_n$ and $b_1,b_2,\ldots b_n$, if $(a_1,a_2,\ldots a_n)\succ(b_1,b_2,\ldots b_n)$, then $\sum_{sym}x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}\ge \sum_{sym}x_1^{b_1}x_2^{b_2}\cdots x_n^{b_n}$

Search up Majorization if you do not know what $\succ$ (which means majorizes) means.

But Muirhead is basically just a generalized AM-GM. Setting the first set of numbers as $1,0,\ldots 0$ and the second set as $\dfrac{1}{n},\dfrac{1}{n},\ldots \dfrac{1}{n}$, we see that $(1,0,\ldots 0)\succ \left(\dfrac{1}{n},\dfrac{1}{n},\ldots \dfrac{1}{n}\right)$ and using Muirhead's gives us the familiar $n$-term AM-GM.

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By AM-GM,

$\dfrac{x^2 + x^2 + y^2 + z^2}{4} \ge \sqrt[4]{x^4y^2z^2}$

so $\dfrac12x^2 + \dfrac14y^2 + \dfrac14z^2 \ge x\sqrt{yz}.$ Similarly, if we duplicate the $y^2$ and $z^2$ terms, we get

$\dfrac14x^2 + \dfrac12y^2 + \dfrac14z^2 \ge y\sqrt{zx},$ and

$\dfrac14x^2 + \dfrac14y^2 + \dfrac12z^2 \ge z\sqrt{xy}.$ Adding these inequalities together, we have

$x^2+y^2+z^2 \ge x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}$ as requested. $\square$

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Ah, you beat me to it with the same solution I posted. Kudos to you on being first.

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