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Let $x,y,z>0$ and $x+y+z=1$. Prove that $$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$$

My work so far:

I use Titu's Lemma:

$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x=$$ $$=\frac{x^4}{xy}+\frac{y^4}{yz}+\frac{z^4}{zx}+\frac{x^2}{\frac zx}+\frac{z^2}{\frac yz}+\frac{y^2}{\frac xy}\ge$$ $$\ge\frac{\left(x^2+y^2+z^2+x+y+z\right)^2}{xy+yz+zx+ \frac zx+\frac yz+\frac xy}=\frac{\left(x^2+y^2+z^2+1\right)^2}{xy+yz+zx+ \frac zx+\frac yz+\frac xy}$$

I need help here.

Community
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Roman83
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  • Could you use Muirhead? The left handside is a symmetric polynomial in $x,y,z$. On the right side you could eliminate the constraint and have a homogeneous expression by replacing $1=(x+y+z)^2$. – Beni Bogosel Aug 01 '16 at 14:07

2 Answers2

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Suppose $x \geq y \geq z$. Using the rearrangements inequality we have $$ \frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x} \geq x^3 \cdot \frac{1}{x}+...+z^3 \cdot \frac{1}{z} =x^2+y^2+z^2 $$ since $x^3,y^3,z^3$ and $1/x,1/y,1/z$ have opposite ordering. The same thing applies to $$ \frac{x^3}{z}+\frac{z^3}{y}+\frac{y^3}{x} \geq x^2+y^2+z^2$$ Thus the left hand side is greater than $2(x^2+y^2+z^2)$. This is immediately greater than the RHS by noting that $1=(x+y+z)^2$.

You could use "Titu's Lemma" if you like, just like in your try, but split in half: $$ \frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x} \geq \frac{(x^2+y^2+z^2)^2}{xy+yz+zx} \geq \frac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2} = x^2+y^2+z^2 $$

Beni Bogosel
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$$S=\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x=\frac{x^4+y^4}{xy}+\frac{y^4+z^4}{yz}+\frac{z^4+x^4}{zx},$$ $$S=\frac{(x^2+y^2)^2-2x^2y^2}{xy}+\frac{(y^2+z^2)^2-2y^2z^2}{yz}+\frac{(z^2+x^2)^2-2z^2x^2}{zx}.$$ Since $(a^2+b^2)^2=(a^2+b^2)(a^2+b^2)\geq 2ab(a^2+b^2)$ by AM-GM, we have $$S\geq 4(x^2+y^2+z^2)-2(xy+yz+zx),$$ $$S\geq\frac{7}{2}(x^2+y^2+z^2)-2(xy+yz+zx)+\frac{x^2+y^2+z^2}{2}.$$

Since $$2(x^2+y^2+z^2)\geq 2(xy+yz+zx)$$ holds (because it is equivalent with $(x-y)^2+(y-z)^2+(z-x)^2$),

we have $$S\geq\frac{3}{2}(x^2+y^2+z^2)+\frac{x^2+y^2+z^2}{2}.$$ Also, $$3(x^2+y^2+z^2)\geq (x+y+z)^2$$ by quadratic-arithmetic inequality. Therefore, $$S\geq\frac{1}{2}(x+y+z)^2+\frac{x^2+y^2+z^2}{2}$$ $$S\geq\frac{1}{2}+\frac{x^2+y^2+z^2}{2}=\frac{x^2+y^2+z^2+1}{2}.$$

alans
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