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If $x,y,z\gt 0$ and $xyz=1$ Then find the minimum value of $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$

$\bf{My\; Try::}$Using Titu's Lemma $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2}\ge 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$

and equality holds when $$x=y=z=1$$

My question is how can we solve it without the above lemma, like using Jensen's Inequality or other inequality.

Please explain me.

Thanks

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    $\begingroup$ ... equality holds when $x=y=z=\color{red}{1}$ isn't it? $\endgroup$ – Surb Jun 27 '16 at 8:03
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This answer shows how Nesbitt's inequality can used with other proof techniques to prove the given inequality.

Nesbitt's inequality

$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geq \frac{3}{2}\tag{$*$}$$


$(1a)$ Observe that

$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)+(x+y+z)=(x+y+z)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\tag{A}$$

Using $(*)$ and (A), we get

$$\sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq \frac{(x+y+z)}{2}\tag{B}$$

Now, using AM-GM inequality, we get

$$\frac{(x+y+z)}{2} \geq \frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\tag{C}$$

Using (B) and (C), we get the desired result.

$(1b)$ Observe that

$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)=(x+y+z)\left[\left(\sum_{cyc}\frac{x}{y+z}\right)-1\right]\tag{D}$$

Using $(*)$ and (C) in (D), we get

$$\sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq 3 \cdot \left(\frac{3}{2}-1\right)=\frac{3}{2}$$


$(2)$

Without loss of generality, let $x \geq y \geq z$

Therefore, $$\frac{x}{y+z} \geq \frac{y}{x+z} \geq \frac{z}{x+y}$$

Now, using the Chebyshev inequality for these two increasing sequences, we get

$$3 \sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq (x+y+z)\sum_{cyc}\left(\frac{x}{y+z}\right)\tag{E}$$

Using $(*)$ and (C) in (E), we get the required result.


$(3)$ Let $f(x)=x^2$

Now, we can write

$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)=\sum_{cyc}\left(\frac{f(x)}{y+z}\right)$$

Note, that $f''(x)=2 > 0$, so the function is convex.

Now, using the Weighted-Jensen inequality, we get

$$\sum_{cyc}\left(\frac{f(x)}{y+z}\right) \geq 3f(M)\tag{F}$$

where using $(*)$, we get

$$3M=\sum_{cyc}\left(\frac{x}{y+z}\right) \geq \frac{3}{2} \Longleftrightarrow M \geq \frac{1}{2}\tag{G}$$

Using (G) in (F), we get the desired result.

Note: Notice that this method is independent of the constraint $xyz=1$


Finally, note that equality is indeed attained when $\boxed{x=y=z=1}$

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Use $$\frac{a^2}{b}\ge 2a-b$$ Then $$\frac{x^2}{y+z}=\frac14\cdot\frac{(2x)^2}{y+z}\ge\frac14(4x-y-z)$$ Similarly $$\frac{y^2}{x+z}\ge\frac14(4y-x-z)$$ $$\frac{z^2}{x+y}\ge\frac14(4z-x-y)$$ Hence

$$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\geq \frac{1}{4}2(x+y+z) = \frac{x+y+z}{2}\geq 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$

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Just find the minimum value of $e=(x+y)(y+z)(z+x)$ because give expression E satisfies $E\ge \frac{3}{[(x+y)(y+z)(z+x)]^{\frac{1}{3}}}$ (using AM-GM inequality). Now you can find the minimum value of $e$ by lagrange multiplier method with constraint $xyz=1$

Hope this helps !

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