# Proving inequality using Cauchy-Schwarz

How to prove $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\leqslant\frac{2(x^2+y^2+z^2)}{x+y+z}$ where $x,y,z$ are all positive real numbers? The hint was to use the Schwarz inequality.

• Maybe you could try the Schwarz inequality instead? ;-) – WimC Mar 31 '13 at 15:45
• Lots of inequality techniques are in Lohwater's unpublished "Introduction to Inequalities" (the link might be bad, look for it in Google if it doesn't work). – vonbrand Mar 31 '13 at 15:50
• Please check the Q. Try $x=2, y=1, z=0$ for e.g. – Macavity Mar 31 '13 at 15:59
• Let $x+y+z = 3$. When $x \to 3^-$, $y,z \to 0^+$, it's clear that LHS is unbounded while RHS is bounded. – user27126 Mar 31 '13 at 16:32
• You must have mistyped the expressions in the inequality because with $x=2, y=1,z=0.1$ the left hand side is greater than the right hand side as Macavity mentioned. On the other hand with $x=y=z=1$ the left hand side is less than the right hand side. – user782220 Apr 10 '13 at 1:22

The reverse inequality holds, and can be shown in the following way:$$~$$ Multiply both sides of the original inequality with an $(x+y+z),$ and use $\dfrac{x^2(x+y+z)}{y+z}=\dfrac{x^3}{y+z}+x^2,$ so that it suffices to check that $$\frac{x^3}{y+z}+\frac{y^3}{z+x}+\frac{z^3}{x+y}\geq x^2+y^2+z^2,$$ Which is a direct application of the Cauchy-Schwarz inequality in accordance with $(xy+yz+zx)\leq(x^2+y^2+z^2)$: $$\sum_{cyc}\frac{x^3}{y+z}\sum_{cyc}x(y+z)\geq (x^2+y^2+z^2)^2,$$ And we are done. Equality holds in the original inequality if and only if $x=y=z,$ or $x=y,z=0$ and its cyclic permutations. $\Box$
• The two sets of terms I used are $\sqrt{\frac{x^3}{y+z}}$ and $\sqrt{x(y+z)}.$ Maybe that form of C-S is better known as "Titu's lemma," since $\frac{x^3}{y+z}=\frac{(x^2)^2}{x(y+z)}$ etc. I think the manipulations are also correct. Please rectify me if I am wrong. – Potla May 7 '13 at 12:57