After expanding we have $$\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{b}+\frac{c}{a}\geq 6+3\sqrt[3]{\left(1-\frac{b}{a}\right)^2 \left(1-\frac{c}{b} \right )^2 \left(1-\frac{a}{c} \right)^2}$$Let $\frac{b}{a}=x, \frac{c}{b}=y$ and $\frac{a}{c}=z$ , we have $xyz=1$ and we need to prove $$\sum \frac{(1-x)^2}{x} \geq 3\sqrt[3]{(1-x)^2(1-y)^2(1-z)^2}$$wich is just AM-GM.

It's equivalent to $\frac{(a-b)^{2}}{ab}+\frac{(b-c)^{2}}{bc}+\frac{(c-a)^{2}}{ca} \geqslant 3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{a^2b^2c^2}}$ which is AM-GM.

Steve12345 wrote: Let $a,b,c$ be positive real numbers. Prove that: $\left (a+b+c \right )\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ) \geq 9+3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{a^2b^2c^2}}$ The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\left (a+b+c \right )\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ) \geq 9+3\sqrt[3]{\frac{2(a-b)^2(b-c)^2(c-a)^2}{a^2b^2c^2}}$$

Steve12345 wrote: Let $a,b,c$ be positive real numbers. Prove that: $\left (a+b+c \right )\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ) \geq 9+3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{a^2b^2c^2}}$ Saudi Arabia JTST -2015

2015 Saudi Arabia JBMO TST 3.4: Let $a,b$ and $c$ be positive numbers with $a^2+b^2+c^2=3$. Prove that $$a+b+c\ge 3\sqrt[5]{abc}$$2015 Saudi Arabia JBMO TST 1.2: Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{\sqrt{(2a+b)(2a+c)}} +\frac{b}{\sqrt{(2b+c)(2b+a)}} +\frac{c}{\sqrt{(2c+a)(2c+b)}} \le 1 $$2015 Saudi Arabia JBMO TST

sqing wrote: 2015 Saudi Arabia JBMO TST 1.2: Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{\sqrt{(2a+b)(2a+c)}} +\frac{b}{\sqrt{(2b+c)(2b+a)}} +\frac{c}{\sqrt{(2c+a)(2c+b)}} \le 1 $$2015 Saudi Arabia JBMO TST 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 or Titu's Lemma, 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

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Kgxtixigct wrote: $$f(a,b,c)\geqslant f(\sqrt{ab},\sqrt{ab},c)\geqslant 0$$ The condition mentioned in the problem is $a^2+b^2+c^2=3$ . However the mixing variable approach you use doesnt preserve this condition and instead preserves $abc= \text { constant}$ . So in my opinion the SMV approach doesn't work @below which part of the comment can you not understand? SMV approaches work only when some particular condition about the variables is kept fixed and the variables are brought closer to each other . But you aren't preserving the condition $a^2+b^2+c^2=3$ with the substitution $(a,b,c) \mapsto (\sqrt{ab},\sqrt{ab},c)$

Aryan-23 wrote: Kgxtixigct wrote: $$f(a,b,c)\geqslant f(\sqrt{ab},\sqrt{ab},c)\geqslant 0$$ The condition mentioned in the problem is $a^2+b^2+c^2=3$ . However the mixing variable approach you use doesnt preserve this condition and instead preserves $abc= \text { constant}$ . So in my opinion the SMV approach doesn't work Could you please explain me in detail that where did i go wrong ? Thank you

Solution of NaPrai: 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*}

Aryan-23 wrote: Kgxtixigct wrote: $$f(a,b,c)\geqslant f(\sqrt{ab},\sqrt{ab},c)\geqslant 0$$ The condition mentioned in the problem is $a^2+b^2+c^2=3$ . However the mixing variable approach you use doesnt preserve this condition and instead preserves $abc= \text { constant}$ . So in my opinion the SMV approach doesn't work @below which part of the comment can you not understand? SMV approaches work only when some particular condition about the variables is kept fixed and the variables are brought closer to each other . But you aren't preserving the condition $a^2+b^2+c^2=3$ with the substitution $(a,b,c) \mapsto (\sqrt{ab},\sqrt{ab},c)$ Thank you.....probably $f\left(\sqrt{\frac{a^2+b^2}{2}},\sqrt{\frac{a^2+b^2}{2}},c\right)$ would work.