crazyfehmy wrote: Show that \[ \sum_{cyc} \sqrt[4]{\frac{(a^2+b^2)(a^2-ab+b^2)}{2}} \leq \frac{2}{3}(a^2+b^2+c^2)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \] for all positive real numbers $a, \: b, \: c.$ First, we have a lemma : with $ x,y > 0 : \frac{x^3+y^3}{2} \le \left(\frac{x^2+y^2}{x+y}\right)^3 $ Proof : Lemma $ \Leftrightarrow (x-y)^4(x^2-xy+y^2) \ge 0 $ ( obviously true). Hence $ \sum\sqrt[4]{\frac{(a^2+b^2)(a^2-ab+b^2)}{2}} = \sum\sqrt[4]{\frac{(a^2+b^2)(a^3+b^3)}{2(a+b)}} \le \sum\frac{a^2+b^2}{a+b} $ Hence we need prove that : $ \sum{\frac{a^2+b^2}{a+b}} \le \frac{2}{3}(a^2+b^2+c^2)\left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{c+a}\right) $ $ \Leftrightarrow \sum\frac{2a^2-b^2-c^2}{b+c} \ge 0 $ Using Chebyshev's inequality : $ \sum\frac{2a^2-b^2-c^2}{b+c} \ge 0 $ Hence we have done.

crazyfehmy wrote: Show that \[ \sum_{cyc} \sqrt[4]{\frac{(a^2+b^2)(a^2-ab+b^2)}{2}} \leq \frac{2}{3}(a^2+b^2+c^2)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \] for all positive real numbers $a, \: b, \: c.$ By HM-GM inequality, we have: $\frac{2}{3}({a^2} + {b^2} + {c^2})\left( {\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}}} \right) \ge \frac{{6({a^2} + {b^2} + {c^2})}}{{2(a + b + c)}} = \frac{{3({a^2} + {b^2} + {c^2})}}{{a + b + c}}$ Use AM-GM and Cauchy-Schwarz ineq, we have: $\sum\limits_{cyc} {\sqrt[4]{{\frac{{({a^2} + {b^2})({a^2} - ab + {b^2})}}{2}}}} \le \sum\limits_{cyc} {\sqrt {\frac{{\frac{{({a^2} + {b^2})}}{2} + ({a^2} - ab + {b^2})}}{2}} } $ $ = \sum\limits_{cyc} {\sqrt {\frac{{3{a^2} - 2ab + 3{b^2}}}{4}} } \le \sqrt {\frac{{3\left( {6({a^2} + {b^2} + {c^2}) - 2(ab + bc + ca)} \right)}}{4}} $ And we need to prove : ${a^2} + {b^2} + {c^2} \ge (a + b + c)\sqrt {\frac{{3({a^2} + {b^2} + {c^2}) - (ab + bc + ca)}}{6}} $ $ \Leftrightarrow {a^2} + {b^2} + {c^2} \ge \frac{{\sqrt {.2{{(a + b + c)}^2}\left( {9({a^2} + {b^2} + {c^2}) - 3(ab + bc + ca)} \right)} }}{6} $ By AM-GM inequlity, we have: $RHS \le \frac{{11({a^2} + {b^2} + {c^2}) + ab + bc + ca}}{{12}} \le {a^2} + {b^2} + {c^2}$ ( by a familiar ineq : $ab+bc+ca \le {a^2} + {b^2} + {c^2}$ ) We have q.d.e Equality holds when $a=b=c$

Is there any solution with jensen ? i was looked like $f(x)=x^{1/4}$

$$(a^2+b^2)(a^2-ab+b^2)=a^4+b^4+2a^2b^2-ab(a^2+b^2)\leq a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4$$$$\sum_{cyc}{\sqrt[4]{\frac{(a^2+b^2)(a^2-ab+b^2)}{2}}}\leq \sum_{cyc}{\sqrt[4]{\frac{a^4+b^4}{2}}}\leq 3\sqrt[4]{a^4+b^4+c^4}\leq \sqrt{3\sqrt{3(a^4+b^4+c^4)}}\geq \frac{3(a^2+b^2+c^2)}{a+b+c}$$How can i tie up the last part ?

ehuseyinyigit wrote: $$(a^2+b^2)(a^2-ab+b^2)=a^4+b^4+2a^2b^2-ab(a^2+b^2)\leq a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4$$$$\sum_{cyc}{\sqrt[4]{\frac{(a^2+b^2)(a^2-ab+b^2)}{2}}}\leq \sum_{cyc}{\sqrt[4]{\frac{a^4+b^4}{2}}}\leq 3\sqrt[4]{a^4+b^4+c^4}\leq \frac{3(a^2+b^2+c^2)}{a+b+c}$$How can i tie up the last part ? Last one is homogenous however using homogenousity doesn't work .

\[LHS=\sum\sqrt[4]{\frac{(a^2+b^2)(a^2-ab+b^2)}{2}}=\sum{\sqrt[4]{\frac{a^2+b^2}{2}.(a^2-ab+b^2)}}\overset{AM-GM}{\leq} \sum{\sqrt{\frac{\frac{a^2+b^2}{2}+a^2-ab+b^2}{2}}}\]\[\sum{\sqrt{3a^2-2ab+3b^2}}\overset{CS}{\leq} \sqrt{(\sum{3a^2-2ab+3b^2})\sum{1}}=\sqrt{3(6\sum{a^2}-2\sum{bc})}\overset{?}{\leq}2RHS\]Also $RHS=\frac{2}{3}\sum{a^2}\sum{\frac{1}{b+c}}\geq \frac{2}{3}\sum{a^2}.\frac{9}{2\sum{a}}=\frac{3\sum{a^2}}{\sum{a}}$. Let's show that $\frac{6\sum{a^2}}{\sum{a}}\overset{?}{\geq} \sqrt{18\sum{a^2}-6\sum{bc}}$. Set $a+b+c=3u,ab+bc+ca=3v^2,abc=w^3$. \[\frac{6(3u^2-2v^2)}{u}=\frac{6(9u^2-6v^2)}{3u}\overset{?}{\geq} \sqrt{18(9u^2-6v^2)-18v^2}=\sqrt{162u^2-126v^2}=3\sqrt{18u^2-14v^2}\]Or \[\frac{2(3u^2-2v^2)}{u}\overset{?}{\geq} \sqrt{18u^2-14v^2}\iff 4(9u^4-12u^2v^2+4v^4)\overset{?}{\geq}18u^4-14u^2v^2\]\[36u^4-48u^2v^2+16v^4\overset{?}{\geq} 18u^4-14u^2v^2\iff 18u^4+16v^4\overset{?}{\geq} 34u^2v^2\]Which is true since $18u^4+16v^4=16u^4+16v^4+2u^4\geq 32u^2v^2+2u^4\geq 34u^2v^2$ as desired.$\blacksquare$