$$\sum\frac{1}{a+b}=\sum\frac{ab\cdot c}{a+b}\leq\sum\frac{\left(\frac{a+b}{2}\right)^2c}{a+b}=\sum\frac{ac+bc}{4}=\frac{ab+bc+ca}{2}\leq\frac{a^2+b^2+c^2}{2}$$

L3435 wrote: $$\sum\frac{1}{a+b}=\sum\frac{ab\cdot c}{a+b}\leq\sum\frac{\left(\frac{a+b}{2}\right)^2c}{a+b}=\sum\frac{ac+bc}{4}=\frac{ab+bc+ca}{2}\leq\frac{a^2+b^2+c^2}{2}$$ A little bit more simple: $$2\sum\frac{1}{a+b}=\sum c\frac{2ab}{a+b}\leq \sum\frac{c(a+b)}{2}=ab+bc+ca\leq a^2+b^2+c^2.$$

gobathegreat wrote: Let $a$, $b$ and $c$ be positive real numbers such that $abc=1$. Prove the inequality: $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \leq \frac{a^2+b^2+c^2}{2}$$ Let $a$, $b$ ,$c$ and $d$ be positive real numbers . Prove the inequality:

$$abc(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} )\leq \frac{a^2}{2}+bc$$$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \leq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2\sqrt{ abc}}\leq \frac{a^2+b^2+c^2}{2abc}$$$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+d}+\frac{1}{d+a} \leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}}{2\sqrt{ abcd}}\leq \frac{a^3+b^3+c^3+d^3}{2abcd}$$$$\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b} \leq \frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+\sqrt[3]{d}}{3\sqrt[3]{ abcd}}\leq \frac{a^3+b^3+c^3+d^3}{3abcd}$$Let $a$, $b$ and $c$ be positive real numbers such that $abc=1$. Prove the inequality: $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \leq \frac{ab+bc+ca}{2}$$$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \leq \frac{a^2}{2}+bc$$

gobathegreat wrote: Let $a$, $b$ and $c$ be positive real numbers such that $abc=1$. Prove the inequality: $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \leq \frac{a^2+b^2+c^2}{2}$$ Zhangyanzong inequalities: Let $a$, $b$ and $c$ be positive real numbers such that $abc=1$. Prove the inequality: $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \leq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\leq \frac{ab+bc+ca}{2}\leq\frac{\sqrt{2}}{4} \left(a\sqrt{b^2+c^2}+b\sqrt{c^2+a^2}+c\sqrt{a^2+b^2}\right)\leq \frac{a^2+b^2+c^2}{2}.$$Let $a_1,a_2,\cdots,a_n$ be positive real numbers such that $a_1 a_2 \cdots a_n=1$. Prove the inequality:

Zhangyanzong inequalities: Let $a_1,a_2,\cdots,a_n$ be positive real numbers such that $a_1 a_2 \cdots a_n=1$. For $k=2,3,\cdots,n-1$ , prove the inequality:

It is even easier just to notice that by AM-GM $\frac{1}{a+b}\leq \frac{1}{2\sqrt{ab}}$ With $abc=1$ We get: $\frac{1}{2\sqrt{ab}}=\frac{1}{2\sqrt{\frac{1}{c}}} =\frac{\sqrt{c}}{2}$ Now we need to prove: $\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\leq \frac{a^2+b^2+c^2}{2}$ By C-S (Engel / Titu): $\frac{a}{\frac{1}{a}}+\frac{b}{\frac{1}{b}}+\frac{c}{\frac{1}{c}} \geq \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\frac{ab+bc+ca}{abc}}$ We now see that from the given $abc=1$ Then by simplifying we get: $(ab+bc+ca)(a^2+b^2+c^2)\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ Or when taking the square root: $\sqrt{(ab+bc+ca)(a^2+b^2+c^2)}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}$ Then from us needing to prove this: $\sqrt{a}+\sqrt{b}+\sqrt{c}\leq a^2+b^2+c^2$ We get from the C-S that we need to prove : $a^2+b^2+c^2\geq \sqrt{ab+bc+ca}\sqrt{a^2+b^2+c^2}$ Which is obvious by dividing both sides by $\sqrt{a^2+b^2+c^2} $ and then squaring : We get $ a^2+b^2+c^2\geq ab+bc+ca $ This is obviously true because of AM-GM: $ \frac{a^2+b^2}{2}\geq ab$ $ \frac{b^2+c^2}{2}\geq bc$ $ \frac{c^2+a^2}{2}\geq ca$ Summing these we get the desired result.

gobathegreat wrote: Let $a$, $b$ and $c$ be positive real numbers such that $abc=1$. Prove the inequality: $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \leq \frac{a^2+b^2+c^2}{2}$$ Zhangyanzong inequalities: Let $a$, $b$ and $c$ be positive real numbers such that $abc=1$. For the positive integer $m$,prove the inequality:

From $AM-GM$ inequality we get that $\frac{ab}{a+b} \leq \frac{a+b}{4}, \frac{bc}{b+c} \leq \frac{b+c}{4}, \frac{ac}{a+c} \leq \frac{a+c}{4}$. We can write our expression like this (using that $abc=1$), so: $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{ac \cdot bc}{ac+bc}+\frac{ab \cdot ac}{ab+ac}+\frac{ab \cdot bc}{ab+bc} \leq \frac{ab+ac+bc}{2}$ Again, by $AM-GM$ inequality we get: $\frac{ab+ac+bc}{2} \leq \frac{a^2+b^2+c^2}{2}$, so $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \leq \frac{a^2+b^2+c^2}{2}$ That is all