Let $x^2=a,y^2=b,z^2=c$. Then we get the following problem: Restated problem wrote: Let $a,b,c \in \mathbb{R^+} \cup \{0\}$. Prove that $$ \frac{b}{2a+1}+\frac{c}{2b+1}+\frac{a}{2c+1} \geq \frac{a}{2a+1}+\frac{b}{2b+1}+\frac{c}{2c+1}$$ WLOG assume that $a \geq b \geq c$. Then we have $\frac{1}{2a+1} \leq \frac{1}{2b+1} \leq \frac{1}{2c+1}$. Thus, By Rearrangement Theorem on these two sequences, we get the desired inequality. Also, Equality holds for $a=b=c$.

gobathegreat wrote: Prove the inequality $$ \frac{y^2-x^2}{2x^2+1}+\frac{z^2-y^2}{2y^2+1}+\frac{x^2-z^2}{2z^2+1} \geq 0$$where $x$, $y$ and $z$ are real numbers $$ \frac{y^2-x^2}{2x^2+1}+\frac{z^2-y^2}{2y^2+1}+\frac{x^2-z^2}{2z^2+1} \geq 0\iff \frac{2y^2+1}{2x^2+1}+\frac{2z^2+1}{2y^2+1}+\frac{2x^2+1}{2z^2+1} \geq 3$$

gobathegreat wrote: Prove the inequality $$ \frac{y^2-x^2}{2x^2+1}+\frac{z^2-y^2}{2y^2+1}+\frac{x^2-z^2}{2z^2+1} \geq 0$$where $x$, $y$ and $z$ are real numbers Prove the inequality $$ \frac{y^2-x^2}{\sqrt{2x^2+1}}+\frac{z^2-y^2}{\sqrt{2y^2+1}}+\frac{x^2-z^2}{\sqrt{2z^2+1}} \geq 0$$where $x$, $y$ and $z$ are real numbers

sqing wrote: Prove the inequality $$ \frac{y^2-x^2}{\sqrt{2x^2+1}}+\frac{z^2-y^2}{\sqrt{2y^2+1}}+\frac{x^2-z^2}{\sqrt{2z^2+1}} \geq 0$$where $x$, $y$ and $z$ are real numbers Solution. The Cauchy-Schwarz's inequality gives that \begin{align*}&\frac{2y^2+1}{\sqrt{2x^2+1}}+\frac{2z^2+1}{\sqrt{2y^2+1}}+\frac{2x^2+1}{\sqrt{2z^2+1}} \ge\frac{\left(\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}\right)^2}{\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}}\\ \Longrightarrow&\frac{2y^2+1}{\sqrt{2x^2+1}}+\frac{2z^2+1}{\sqrt{2y^2+1}}+\frac{2x^2+1}{\sqrt{2z^2+1}} \ge\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}\\ \Longrightarrow&\frac{2y^2-2x^2}{\sqrt{2x^2+1}}+\frac{2z^2-2y^2}{\sqrt{2y^2+1}}+\frac{2x^2-2z^2}{\sqrt{2z^2+1}} \ge0\\ \Longrightarrow&\frac{y^2-x^2}{\sqrt{2x^2+1}}+\frac{z^2-y^2}{\sqrt{2y^2+1}}+\frac{x^2-z^2}{\sqrt{2z^2+1}} \geq 0. \end{align*}As desired. $\blacksquare$

ytChen wrote: sqing wrote: Prove the inequality $$ \frac{y^2-x^2}{\sqrt{2x^2+1}}+\frac{z^2-y^2}{\sqrt{2y^2+1}}+\frac{x^2-z^2}{\sqrt{2z^2+1}} \geq 0$$where $x$, $y$ and $z$ are real numbers Solution. The Cauchy-Schwarz's inequality gives that \begin{align*}&\frac{2y^2+1}{\sqrt{2x^2+1}}+\frac{2z^2+1}{\sqrt{2y^2+1}}+\frac{2x^2+1}{\sqrt{2z^2+1}} \ge\frac{\left(\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}\right)^2}{\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}}\\ \Longrightarrow&\frac{2y^2+1}{\sqrt{2x^2+1}}+\frac{2z^2+1}{\sqrt{2y^2+1}}+\frac{2x^2+1}{\sqrt{2z^2+1}} \ge\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}\\ \Longrightarrow&\frac{2y^2-2x^2}{\sqrt{2x^2+1}}+\frac{2z^2-2y^2}{\sqrt{2y^2+1}}+\frac{2x^2-2z^2}{\sqrt{2z^2+1}} \ge0\\ \Longrightarrow&\frac{y^2-x^2}{\sqrt{2x^2+1}}+\frac{z^2-y^2}{\sqrt{2y^2+1}}+\frac{x^2-z^2}{\sqrt{2z^2+1}} \geq 0. \end{align*}As desired. $\blacksquare$ $$\frac{2y^2+1}{\sqrt{2x^2+1}}+\frac{2z^2+1}{\sqrt{2y^2+1}}+\frac{2x^2+1}{\sqrt{2z^2+1}} \ge\frac{\left(\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}\right)^2}{\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}}=\sqrt{2x^2+1}+\sqrt{2y^2+1}+\sqrt{2z^2+1}\iff\frac{y^2-x^2}{\sqrt{2x^2+1}}+\frac{z^2-y^2}{\sqrt{2y^2+1}}+\frac{x^2-z^2}{\sqrt{2z^2+1}} \geq 0.$$Thanks.

sqing wrote: gobathegreat wrote: Prove the inequality $$ \frac{y^2-x^2}{2x^2+1}+\frac{z^2-y^2}{2y^2+1}+\frac{x^2-z^2}{2z^2+1} \geq 0$$where $x$, $y$ and $z$ are real numbers $$ \frac{y^2-x^2}{2x^2+1}+\frac{z^2-y^2}{2y^2+1}+\frac{x^2-z^2}{2z^2+1} \geq 0\iff \frac{2y^2+1}{2x^2+1}+\frac{2z^2+1}{2y^2+1}+\frac{2x^2+1}{2z^2+1} \geq 3$$ Follows from Rearrangement inequality.

gobathegreat wrote: Prove the inequality $$ \frac{y^2-x^2}{2x^2+1}+\frac{z^2-y^2}{2y^2+1}+\frac{x^2-z^2}{2z^2+1} \geq 0$$where $x$, $y$ and $z$ are real numbers Ignore......

AlastorMoody wrote: gobathegreat wrote: Prove the inequality $$ \frac{y^2-x^2}{2x^2+1}+\frac{z^2-y^2}{2y^2+1}+\frac{x^2-z^2}{2z^2+1} \geq 0$$where $x$, $y$ and $z$ are real numbers We know by Titu' Lemma, $$\sum_{cyc} \frac {y^2}{2x^2+1} \geq \frac {(x+y+z)^2}{3+2\sum x^2} $$ And, $$\sum_{cyc} \frac {x^2}{2x^2+1} \geq \frac {(x+y+z)^2}{3+2\sum x^2} $$ Hence combining both the above inequalities, $$\text {LHS} \geq 0$$Hence, Proved This doesn't prove the given inequality. You can't just subtract inequalities in your solution (cause the sign of the second inequality will change).

gobathegreat wrote: Prove the inequality $$ \frac{y^2-x^2}{2x^2+1}+\frac{z^2-y^2}{2y^2+1}+\frac{x^2-z^2}{2z^2+1} \geq 0$$where $x$, $y$ and $z$ are real numbers Greece 2005 here