WakeUp wrote: Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Prove that \[\sqrt[4]{\frac{\sqrt{3}}{a}+6\sqrt{3}b}+\sqrt[4]{\frac{\sqrt{3}}{b}+6\sqrt{3}c}+\sqrt[4]{\frac{\sqrt{3}}{c}+6\sqrt{3}a}\le\frac{1}{abc}\] When does inequality hold? Denote , \[S =\sqrt[4]{\frac{\sqrt{3}}{a}+6\sqrt{3}b}+\sqrt[4]{\frac{\sqrt{3}}{b}+6\sqrt{3}c}+\sqrt[4]{\frac{\sqrt{3}}{c}+6\sqrt{3}a}\] Note that , By holder inequality \[ \left(\frac{\sqrt{3}}{a}+6\sqrt{3}b+\frac{\sqrt{3}}{b}+6\sqrt{3}c+\frac{\sqrt{3}}{c}+6\sqrt{3}a \right)(1+1+1)^3 \ge S^4 \] Thus it suffice to show that , \[ 27\left(\frac{\sqrt{3}}{a}+6\sqrt{3}b+\frac{\sqrt{3}}{b}+6\sqrt{3}c+\frac{\sqrt{3}}{c}+6\sqrt{3}a \right) \le \frac{1}{a^4b^4c^4} \] Now again note that , since $\frac{1}{3\sqrt{3}} \ge abc \implies \frac{1}{3\sqrt{3}a^3b^3c^3} \ge 27$ Thus , It suffice to show that, \[ \frac{1}{3\sqrt{3}a^3b^3c^3}\left(\frac{\sqrt{3}}{a}+6\sqrt{3}b+\frac{\sqrt{3}}{b}+6\sqrt{3}c+\frac{\sqrt{3}}{c}+6\sqrt{3}a \right) \le \frac{1}{a^4b^4c^4} \] Which is equivalent to, \[ \frac{1}{3} \left(\frac{1}{a}+6b+\frac{1}{b}+6c+\frac{1}{c}+6a \right) \le \frac{1}{abc} \] Which is true since , \[\frac{1}{3} \left(\frac{1}{a}+6b+\frac{1}{b}+6c+\frac{1}{c}+6a \right) = \frac{1}{3abc}(1+6abc(a+b+c)) \] And since $(ab+bc+ac)^2= a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c) \ge 3abc(a+b+c) $ Thus, \[ \frac{1}{3abc}(1+6abc(a+b+c)) \le \frac{1}{3abc}(1+2)= \frac{1}{abc} \Box \]

We apply Jensen's Inequality on the concave function $f(x)=x^{1/4}$. Then, we get $LHS \le \frac{3}{3^8} \left( \sqrt[4]{6(a+b+c)+ \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)} \right)$ Since $ab+bc+ca=1$, we have $\sum_{cyc} \frac{1}{a} = \frac{1}{abc}$. Also, $1=(ab+bc+ca)^2 \ge 3abc(a+b+c) \implies a+b+c \le \frac{1}{3abc}$. Again, using AM-GM inequality on the given condition, we have $abc \le \frac{1}{3\sqrt{3}}$. Then, we get $LHS \le 3^{7/8}\sqrt[4]{\frac{3}{abc}} \le \frac{1}{abc}$. QED.

WakeUp wrote: Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Prove that \[\sqrt[4]{\frac{\sqrt{3}}{a}+6\sqrt{3}b}+\sqrt[4]{\frac{\sqrt{3}}{b}+6\sqrt{3}c}+\sqrt[4]{\frac{\sqrt{3}}{c}+6\sqrt{3}a}\le\frac{1}{abc}\] When does inequality hold? http://www.artofproblemsolving.com/Forum/viewtopic.php?p=191113&sid=2206b79f7b3349f1b83e5ee2581507c4#p191113

The problem is equivalent to Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$. Prove that\[\sqrt[4]{\frac{1}{a}+2b}+\sqrt[4]{\frac{1}{b}+2c}+\sqrt[4]{\frac{1}{c}+2a}\le\frac{3\sqrt[4]{3}}{abc} .\]

WakeUp wrote: Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Prove that \[\sqrt[4]{\frac{\sqrt{3}}{a}+6\sqrt{3}b}+\sqrt[4]{\frac{\sqrt{3}}{b}+6\sqrt{3}c}+\sqrt[4]{\frac{\sqrt{3}}{c}+6\sqrt{3}a}\le\frac{1}{abc}\]When does inequality hold? Let $\sqrt[4]{\frac{\sqrt{3}}{a}+6\sqrt{3}b}+\sqrt[4]{\frac{\sqrt{3}}{b}+6\sqrt{3}c}+\sqrt[4]{\frac{\sqrt{3}}{c}+6\sqrt{3}a}$$=$$T$ By Power mean inequality $T$$\le$$3$$×$$\sqrt[4]{\frac{\sum_{cyc}{\frac{\sqrt{3}}{a}+6×\sqrt{3}×b}}{3}}\le\frac{1}{abc}$ multiply both sides of the inequality by $\sqrt[4]{abc}$ And we know that $abc(a+b+c)$$\le$$\frac{1}{3}×(ab+bc+ac)^2$$=$$\frac{1}{3}$ then let's we show that $3×\sqrt[4]{\sqrt{3}}\le\frac{1}{abc}×\sqrt[4]{abc}$ Besides $a×b×c\le\sqrt{(\frac{ab+ac+bc}{3})^3}=\frac{1}{3×\sqrt{3}}$ we use this and we are done.

WakeUp wrote: Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Prove that \[\sqrt[4]{\frac{\sqrt{3}}{a}+6\sqrt{3}b}+\sqrt[4]{\frac{\sqrt{3}}{b}+6\sqrt{3}c}+\sqrt[4]{\frac{\sqrt{3}}{c}+6\sqrt{3}a}\le\frac{1}{abc}\]When does inequality hold? Should be 'when does equality hold' right?

WakeUp wrote: Now again note that , since $\frac{1}{3\sqrt{3}} \ge abc \implies \frac{1}{3\sqrt{3}a^3b^3c^3} \ge 27$ i dont get this one,can anyone please explain?