Cauchy-Schwarz only

parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that $\frac{y^2 + z^2}{x}+\frac{z^2 + x^2}{y}+\frac{x^2 + y^2}{z}\ge 2(x + y + z)$. $$\bigg( \frac{y^2}{x}+\frac{z^2 }{y}+\frac{x^2 }{z} +\frac{z^2}{x}+\frac{x^2}{y}+\frac{ y^2}{z}\bigg) (2x+2y+2z) \geq 4(x+y+z)^2$$$$\frac{y^2 + z^2}{x}+\frac{z^2 + x^2}{y}+\frac{x^2 + y^2}{z}\ge 2(x + y + z)$$

parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that $\frac{y^2 + z^2}{x}+\frac{z^2 + x^2}{y}+\frac{x^2 + y^2}{z}\ge 2(x + y + z)$. $$\sum_{cyc}\frac{y^2+z^2}{x}=\sum_{cyc}\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\geq\sum_{cyc}\frac{(x+y)^2}{x+y}=2(x+y+z).$$

Let $x, y, z$ be positive real numbers such that $x+y+z=3$. Prove that $$\sum_{cyc}\frac{y^2+z^2}{x} \geq 3(x^2+y^2+z^2-1)$$.

parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that $\frac{y^2 + z^2}{x}+\frac{z^2 + x^2}{y}+\frac{x^2 + y^2}{z}\ge 2(x + y + z)$. Using AM-GM, $$ LHS = (x^2+y^2+z^2)\sum\frac{1}{x} - \sum x \geq(xy+yz+zx)\sum\frac{1}{x} - \sum x$$$$ \sum x+\sum_{cyc}\frac{yz}{x}\geq2(x+y+z)$$

khan.academy wrote: Let $x, y, z$ be positive real numbers such that $x+y+z=3$. Prove that $$\sum_{cyc}\frac{y^2+z^2}{x} \geq 3(x^2+y^2+z^2-1)$$. the first step trivialises the problem $$(x^2+y^2+z^2)\sum\frac{1}{x} - \sum x \geq 3\sum x^2 -3 $$It remains to prove that $$\sum \frac {1}{x} \geq 3$$which is easy be AM-GM.

khan.academy wrote: Let $x, y, z$ be positive real numbers such that $x+y+z=3$. Prove that $$\sum_{cyc}\frac{y^2+z^2}{x} \geq 3(x^2+y^2+z^2-1)$$. Solution: Rearraging and using AM-HM we get, $$\sum_{cyc} \frac{x^2+y^2+z^2}{x} \geq \frac{9(x^2+y^2+z^2)}{(x+y+z)} \implies \sum_{cyc}\frac{y^2+z^2}{x} \geq 3(x^2+y^2+z^2-1)$$

parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that $\frac{y^2 + z^2}{x}+\frac{z^2 + x^2}{y}+\frac{x^2 + y^2}{z}\ge 2(x + y + z)$. By AM-GM, $$\frac{y^2}{x}+x\geq 2y, \\ \frac{ z^2}{x}+x\geq 2z,...$$

http://olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/09/crmo-14-2.pdf http://olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/09/sol-crmo-14-2.pdf

This is direct from Titu's Lemma (also known as Engel form of CS inequality).

parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that $\frac{y^2 + z^2}{x}+\frac{z^2 + x^2}{y}+\frac{x^2 + y^2}{z}\ge 2(x + y + z)$. By Titu' Lemma $$ \text {LHS } \geq \frac {4 (x+y+z)^2}{2 (x+y+z)} =2 (x+y+z) $$

parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that $\frac{y^2 + z^2}{x}+\frac{z^2 + x^2}{y}+\frac{x^2 + y^2}{z}\ge 2(x + y + z)$. We have: $LHS\geq 2(\frac{yz }{x} + \frac{zx}{y}+\frac{xy}{z})$ $=(\frac{yz}{x}+\frac{zx}{y})+(\frac{zx}{y}+\frac{xy}{z})+(\frac{xy}{z}+\frac{yz}{x})$ $\geq 2(x+y+z) = RHS$ (use AM-GM). So we have $Q.E.D$. And equality occurs when $x = y = z$

SBM wrote: parmenides51 wrote: Let $x, y, z$ be positive real numbers. Prove that $\frac{y^2 + z^2}{x}+\frac{z^2 + x^2}{y}+\frac{x^2 + y^2}{z}\ge 2(x + y + z)$. We have: $LHS\geq 2(\frac{yz }{x} + \frac{zx}{y}+\frac{xy}{z})$ $=(\frac{yz}{x}+\frac{zx}{y})+(\frac{zx}{y}+\frac{xy}{z})+(\frac{xy}{z}+\frac{yz}{x})$ $\geq 2(x+y+z) = RHS$ (use AM-GM). So we have $Q.E.D$. And equality occurs when $x = y = z$ Titu usage kills it.