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For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$ $$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$

My solution$:$

Let $x=a^2,\,y=b^2,\,z=c^2$ then $a+b+c=1,$ we need to prove$:$ $$\sum\limits_{cyc} \frac{a^4+b^2 c^2}{a^2 \sqrt{2(b^2+c^2)}} \geqq 1\Leftrightarrow \sum\limits_{cyc} \frac{a^4+b^2 c^2}{a^2 \sqrt{2(b^2+c^2)}} \geqq a+b+c$$

By AM-GM$:$ $$\text{LHS} = \sum\limits_{cyc} \frac{a^4+b^2 c^2}{a \sqrt{2a^2(b^2+c^2)}} \geqq \sum\limits_{cyc} \frac{2(a^4+b^2c^2)}{a(2a^2+b^2+c^2)} \geqq a+b+c$$

Last inequality is true by SOS$:$

$$\sum\limits_{cyc} \frac{2(a^4+b^2c^2)}{a(2a^2+b^2+c^2)}-a-b-c=\sum\limits_{cyc} {\frac {{c}^{2} \left( a-b \right) ^{2} \left( a+b \right) \left( 2\, {a}^{2}+ab+2\,{b}^{2}+{c}^{2} \right) }{a \left( 2\,{a}^{2}+{b}^{2}+{c }^{2} \right) b \left( {a}^{2}+2\,{b}^{2}+{c}^{2} \right) }} \geqq 0$$

PS:Is there any another solution for original inequality or the last inequality$?$

Thank you!

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  • $\begingroup$ You may use the inequality $1\geq (x+y+z) \geq \frac{1}{3}$ to avoid $4$th power terms. $\endgroup$
    – Alapan Das
    May 21, 2020 at 1:48
  • $\begingroup$ @AlapanDas I tried, but I don't know what to do next? $\endgroup$
    – NKellira
    May 21, 2020 at 2:03

2 Answers 2

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Let $x\geq y\geq z$.

Thus, by C-S we obtain: $$\sum_{cyc}\frac{x^2+yz}{\sqrt{2x^2(y+z)}}=\sum_{cyc}\frac{x^2-xy-xz+yz}{\sqrt{2x^2(y+z)}}+\sum_{cyc}\frac{xy+xz}{\sqrt{2x^2(y+z)}}=$$ $$=\sum_{cyc}\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\sum_{cyc}\frac{\sqrt{2(y+z)}}{2}\geq$$ $$\geq \frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\frac{(y-x)(y-z)}{\sqrt{2y^2(x+z)}}+\frac{1}{2}\sum_{cyc}(\sqrt{y}+\sqrt{z})=$$ $$=\frac{x-y}{\sqrt2}\left(\frac{x-z}{x\sqrt{y+z}}-\frac{y-z}{y\sqrt{x+z}}\right)+1\geq$$ $$\geq \frac{x-y}{\sqrt2}\left(\frac{\frac{x}{y}(y-z)}{x\sqrt{y+z}}-\frac{y-z}{y\sqrt{x+z}}\right)+1=$$ $$=\frac{(x-y)(y-z)}{y\sqrt2}\left(\frac{1}{\sqrt{y+z}}-\frac{1}{\sqrt{x+z}}\right)+1\geq1.$$

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Another way.

Since $$\left(yz-\frac{1}{2}xy-\frac{1}{2}zx, zx-\frac{1}{2}yz-\frac{1}{2}xy,xy-\frac{1}{2}zx-\frac{1}{2}yz\right)$$ and $$\left(\frac{1}{x\sqrt{2(y+z)}},\frac{1}{y\sqrt{2(z+x)}},\frac{1}{z\sqrt{2(x+y)}}\right)$$ have the same ordering, by AM-GM and Chebyshov we obtain: $$\sum_{cyc}\frac{x^2+yz}{x\sqrt{2(y+z)}}-1=\sum_{cyc}\left(\frac{x^2+yz}{x\sqrt{2(y+z)}}-\sqrt{x}\right)=$$ $$=\sum_{cyc}\frac{x^2+yz-x\sqrt{2x(y+z)}}{x\sqrt{2(y+z)}}\geq \sum_{cyc}\frac{x^2+yz-\frac{1}{2}x(2x+y+z)}{x\sqrt{2(y+z)}}=$$ $$=\sum_{cyc}\frac{yz-\frac{1}{2}xy-\frac{1}{2}zx}{x\sqrt{2(y+z)}}\geq\frac{1}{3}\sum_{cyc}\left(yz-\frac{1}{2}xy-\frac{1}{2}zx\right)\sum_{cyc}\frac{1}{x\sqrt{2(y+z)}}=0.$$

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