Let $x, y, z>0$. Prove that: $$\frac{x^3}{2x^2+y^2}+\frac{y^3}{2y^2+z^2}+\frac{z^3}{2z^2+x^2}\ge\frac{x+y+z}{3}$$
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How does $a,b,c$ relate to $x,y,z$? – S.D. Aug 16 '13 at 10:54
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where is a,b,c my friend – Shobhit Aug 16 '13 at 10:56
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Oh... I was wrong.I edited it. Thank you so much! :) – my_melody Aug 16 '13 at 10:58
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Are you absolutely sure of those fractions? They are not symmetric, which is unusual. – L. F. Aug 16 '13 at 11:01
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@KhủngLongBạoChúa would u check the question L.F. has a point. – Shobhit Aug 16 '13 at 11:04
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Sorry. I was wrong again. Thanks :) . I edited . – my_melody Aug 16 '13 at 11:08
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I don't know if it's useful, but setting $\frac{x^3}{2x^2+y^2}=\frac x2-\frac{xy^2}{4x^2+2y^2}$ may gives some idees for other approaches. – Bart Michels Aug 16 '13 at 12:01
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Thanks. But tt's not simple to solve:( ... So I posted it to ask people for help :-< I think we should SOS to solve this problem :-< – my_melody Aug 16 '13 at 17:26
2 Answers
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I have a direction which may be fruitful but didn't work out all cases. Rewrite the inequality as: $$ \frac{x^3}{2x^2+y^2}-\frac{x}{3}+\frac{y^3}{2y^2+z^2}-\frac{y}{3}+\frac{z^3}{2z^2+x^2}-\frac{z}{3}\ge0 $$
Let $f(t)=\frac{1-t^2}{2+t^2}$, then the inequality is equivalent to: $xf(a)+yf(b)+zf(c)\ge0$, where $a=y/x, b=z/y $ and $c=x/z$.
Case 1. Assume that $a,b,c \in (0,2]$. Note that $f(t)\ge \frac{1-t}{2}$ for each $t \in (0,2]$. Then we have $$xf(a)+yf(b)+zf(c) \ge \frac{x}{2}(1-\frac{y}{x})+\frac{y}{2}(1-\frac{z}{y})+\frac{z}{2}(1-\frac{x}{z})=0$$
Case 1 is settled.
Case 2 $x\ge y\ge z$ and $x\ge 2z$ remains to be settled.
Case 3 $z\ge y\ge x$ and $z\ge 2y$ OR $y\ge 2x$ remains to be settled.
ivan
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