Stronger: If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that $$\frac{abc}{3\sqrt{2}\sqrt[4]{27}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$$

Dadgarnia wrote: If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that $$\frac{abc}{3\sqrt{2}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$$ I think this inequality is not a nice problem in Iran TST. We can prove the stronger inequality of socrates by AM-GM: $$\frac{abc}{3\sqrt{2}\sqrt[4]{27}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$$ Indeed, since $$\begin{aligned} \dfrac{1}{abc} \cdot \sum_{cyc} \dfrac{a}{a^2+1} & =\sum_{cyc}\dfrac{1}{a^2bc+bc}=\sum_{cyc}\dfrac{1}{a(a+b+c)+bc} =\sum_{cyc}\dfrac{1}{(a+b)(a+c)} \\ & =\dfrac{2(a+b+c)}{(a+b)(b+c)(c+a)} \leq \dfrac{2abc}{8abc}=\dfrac{1}{4}. \end{aligned}$$ It suffices to show that $$ \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq \dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$$ Notice that $$ (a+b)(b+c)(c+a) \geq \dfrac{8(a+b+c)(ab+bc+ca)}{9} \geq \dfrac{8(a+b+c)\sqrt{3abc(a+b+c)}}{9}=\dfrac{8\sqrt{3} a^2b^2c^2}{9}.$$ Hence, $$ (a^3+b^3)(b^3+c^3)(c^3+a^3) \geq ab(a+b)\cdot bc(b+c) \cdot ca(c+a) \geq \dfrac{8\sqrt{3}a^4b^4c^4}{9}.$$ Now, using the AM-GM inequality, we have $$ (ab+1)(bc+1)(ca+1)=\dfrac{(abc+a)(abc+b)(abc+c) }{abc}\leq \dfrac{1}{abc}\left(\dfrac{abc+a+abc+b+abc+c}{3}\right)^3=\dfrac{64a^2b^2c^2}{27}.$$ Therefore, $$ \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq 3\sqrt[3]{\dfrac{\sqrt{(a^3+b^3)(b^3+c^3)(c^3+a^3)}}{(ab+1)(bc+1)(ca+1)}} \geq 3\sqrt[3]{\dfrac{\sqrt{\dfrac{8\sqrt{3} a^4b^4c^4}{9}}}{\dfrac{64a^4b^4c^4}{27}}}=\dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$$ The proof is completed. Equality holds for $a=b=c=\sqrt{3}.$

The stronger inequality If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that$$\frac{1}{\sqrt[4]{12}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}.$$

sqing wrote: The stronger inequality If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that$$\frac{1}{\sqrt[4]{12}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}.$$ It follows from $$ \dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}=\dfrac{2a^2b^2c^2}{(a+b)(b+c)(c+a)} \leq \dfrac{2a^2b^2c^2}{\dfrac{8\sqrt{3} a^2b^2c^2}{9}}=\dfrac{9}{4\sqrt{3}}.$$ $$\sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq \dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$$

quykhtn-qa1 wrote: Dadgarnia wrote: If $a,b,c$ are positive real numbers such that $a+b+c=abc$ prove that $$\frac{abc}{3\sqrt{2}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$$ I think this inequality is not a nice problem in Iran TST. We can prove the stronger inequality of socrates by AM-GM: $$\frac{abc}{3\sqrt{2}\sqrt[4]{27}}\left ( \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \right )\geq \sum_{cyc}\frac{a}{a^2+1}$$ Indeed, since $$\begin{aligned} \dfrac{1}{abc} \cdot \sum_{cyc} \dfrac{a}{a^2+1} & =\sum_{cyc}\dfrac{1}{a^2bc+bc}=\sum_{cyc}\dfrac{1}{a(a+b+c)+bc} =\sum_{cyc}\dfrac{1}{(a+b)(a+c)} \\ & =\dfrac{2(a+b+c)}{(a+b)(b+c)(c+a)} \leq \dfrac{2abc}{8abc}=\dfrac{1}{4}. \end{aligned}$$ It suffices to show that $$ \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq \dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$$ Notice that $$ (a+b)(b+c)(c+a) \geq \dfrac{8(a+b+c)(ab+bc+ca)}{9} \geq \dfrac{8(a+b+c)\sqrt{3abc(a+b+c)}}{9}=\dfrac{8\sqrt{3} a^2b^2c^2}{9}.$$ Hence, $$ (a^3+b^3)(b^3+c^3)(c^3+a^3) \geq ab(a+b)\cdot bc(b+c) \cdot ca(c+a) \geq \dfrac{8\sqrt{3}a^4b^4c^4}{9}.$$ Now, using the AM-GM inequality, we have $$ (ab+1)(bc+1)(ca+1)=\dfrac{(abc+a)(abc+b)(abc+c) }{abc}\leq \dfrac{1}{abc}\left(\dfrac{abc+a+abc+b+abc+c}{3}\right)^3=\dfrac{64a^2b^2c^2}{27}.$$ Therefore, $$ \sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq 3\sqrt[3]{\dfrac{\sqrt{(a^3+b^3)(b^3+c^3)(c^3+a^3)}}{(ab+1)(bc+1)(ca+1)}} \geq 3\sqrt[3]{\dfrac{\sqrt{\dfrac{8\sqrt{3} a^4b^4c^4}{9}}}{\dfrac{64a^4b^4c^4}{27}}}=\dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$$ The proof is completed. Equality holds for $a=b=c=\sqrt{3}.$ Your proof is truly amazing! What would be some motivations for this solution? Thanks!

After homogenizing the given inequality is equivalent to : $\sqrt[4]{A}(\sum{\frac{\sqrt{a^3+b^3}}{ab(A+c)}})$ $\ge$ $\frac{6\sqrt{2}\sqrt[4]{(abc)^3}}{(a+b)(b+c)(c+a)}$ , where $A=a+b+c$ Now by AM-GM we have that : $LHS$ $\ge$ $\frac{3\sqrt{2}\sqrt[4]{27(abc)^3}}{\sqrt[3]{(abc)^2(A+a)(A+b)(A+c)}}$ Thus we must show that : $\frac{\sqrt[4]{27}}{\sqrt[3]{(abc)^2 \prod(A+a)}}$ $\ge$ $\frac{2}{(a+b)(b+c)(c+a)}$ But we'll prove the stronger inequality : $(a+b)(b+c)(c+a)$ $\ge$ $2\sqrt[3]{(abc)^2\prod{(A+a)}}$ Again using AM-GM $LHS=\frac{\sum{(Aa+bc)(b+c)}}{3}$ $\ge$ $2\sqrt[3]{(abc)\prod{(Aa+bc)}}$ So it suffices to show that $\prod{(Aa+bc)}$ $\ge$ $(abc) \prod{(A+a)}$ After expending it remains to prove that $A^2((ab)^2+(bc)^2+(ac)^2)+Aabc(a^2+b^2+c^2)$ $\ge$ $A^2(abc)(a+b+c)+A(abc)(ab+bc+ac)$ wich is obviously true .