Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{\sqrt{(2a+b)(2a+c)}} +\frac{b}{\sqrt{(2b+c)(2b+a)}} +\frac{c}{\sqrt{(2c+a)(2c+b)}} \leq 1$$$$\leq \sqrt{\frac{bc}{(2a+b)(2a+c)}}+\sqrt{\frac{ca}{(2b+c)(2b+a)}}+\sqrt{\frac{ab}{(2c+a)(2c+b)}}\leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}} $$$$\sqrt{\frac{a^2+bc}{(2a+b)(2a+c)}}+\sqrt{\frac{b^2+ca}{(2b+c)(2b+a)}}+\sqrt{\frac{c^2+ab}{(2c+a)(2c+b)}}\ge \sqrt{2}$$

By Cauchy-Schwarz's inequality, we have \begin{align*}\sum \frac{a}{\sqrt{(2a+b)(2a+c)}} &= \sum \frac{a}{\sqrt{(a+b+a)(a+a+c)}} \\&\le \sum \frac{a}{a+\sqrt{ab}+\sqrt{ac}} \\&= \sum \frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} \\&= 1\end{align*}

parmenides51 wrote: Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{\sqrt{(2a+b)(2a+c)}} +\frac{b}{\sqrt{(2b+c)(2b+a)}} +\frac{c}{\sqrt{(2c+a)(2c+b)}} \le 1 $$ sqing wrote: Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{\sqrt{(2a+b)(2a+c)}} +\frac{b}{\sqrt{(2b+c)(2b+a)}} +\frac{c}{\sqrt{(2c+a)(2c+b)}} \leq 1$$ By AM-GM, we have, $$\sum_{cyc}\left(\frac{a}{\sqrt{(2a+b)(2a+c)}}\right)=\sum_{cyc}\left(a\cdot\sqrt{\frac{1}{(2a+b)}\cdot\frac{1}{(2a+c)}}\right)\leqslant \sum_{cyc}\frac{a}{2}\cdot\left(\frac{1}{2a+b}+\frac{1}{2a+c}\right).$$So, it suffice to prove that, $$\sum_{cyc}\frac{a}{2}\cdot\left(\frac{1}{2a+b}+\frac{1}{2a+c}\right)\leqslant 1$$or $$\sum_{cyc}\left(\frac{a}{2a+b}+\frac{a}{2a+c}\right)\leqslant 2$$or$$\sum_{cyc}\left(2+\frac{2a}{2a+b}-1+\frac{2a}{2a+c}-1\right)\leqslant 4$$or$$\sum_{cyc}\left(2-\frac{b}{2a+b}-\frac{c}{2a+c}\right)\leqslant 4$$or $$\sum_{cyc}\frac{b}{2a+b}+\sum_{cyc}\frac{c}{2a+c}\geqslant 2$$or$$\sum_{cyc}\frac{b^2}{2ab+b^2}+\sum_{cyc}\frac{c^2}{2ac+c^2}\geqslant 2.$$By Angle form of Cauchy-Schwarz's inequality, we have, $$\sum_{cyc}\frac{b^2}{2ab+b^2}+\sum_{cyc}\frac{c^2}{2ac+c^2}\geqslant \frac{(a+b+c)^2}{b^2+2ab+c^2+2bc+a^2+2ac}+\frac{(a+b+c)^2}{a^2+2ab+b^2+2bc+c^2+2ca}=2\frac{(a+b+c)^2}{(a+b+c)^2}=2$$which was desired. $\blacksquare$ Thank You, Best Regards Aditya Raj

Let $a,b,c>0$ .Prove that $$\frac{1}{(ka+b)(ka+c)}+\frac{1}{(kb+c)(kb+a)}+\frac{1}{(kc+a)(kc+b)}\ge \frac {9}{(k + 1)^{2}(ab + bc + ca)}.$$Where $ k\ge 2.$ Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{\sqrt{b^2+c^2}} +\frac{b}{\sqrt{c^2+a^2}} +\frac{c}{\sqrt{a^2+b^2}} \geq\frac{3\sqrt{2}}{2} \cdot \frac{ab+bc+ca} {a^2+b^2+c^2}$$

parmenides51 wrote: Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{\sqrt{(2a+b)(2a+c)}} +\frac{b}{\sqrt{(2b+c)(2b+a)}} +\frac{c}{\sqrt{(2c+a)(2c+b)}} \le 1 $$ T6/540. Let $(x ; y ; z)$ be a permutation of $3$ positive numbers $a, b, c$. Show that $$\frac{x^{2}}{\sqrt{a b(a+2 c)(b+2 c)}}+\frac{y^{2}}{\sqrt{b c(b+2 a)(c+2 a)}} +\frac{z^{2}}{\sqrt{c a(c+2 b)(a+2 b)}} \geq 1.$$

Let $a,b,c$ be positive real numbers. Prove that $$\frac{b+c}{\sqrt{(2a+b)(2a+c)}} +\frac{c+a}{\sqrt{(2b+c)(2b+a)}} +\frac{a+b}{\sqrt{(2c+a)(2c+b)}} \geq 2 $$