Let $a, b>0$. Prove that$$ (a^3+b^3+a^3b^3)(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} ) +36\ge \frac{15}{2}(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}) $$

Hyperabola wrote: Let $a, b>0$. Prove that:$$ (a^3+b^3+a^3b^3)(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} ) +27 \ge 6(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}) $$ Obvious by $x^3 +\frac 1{x^3}+10\geq 6\left(x+\frac 1x\right).$ Full Solution $$ (a^3+b^3+a^3b^3)(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} ) +27 \ge 6(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}) $$$$\Leftrightarrow \frac{a^3}{b^3}+\frac{b^3}{a^3}+a^3+\frac{1}{a^3}+b^3+\frac 1{b^3}+30\geq 6\left(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}\right)$$We only need to prove $x>0,x^3 +\frac 1{x^3}+10\geq 6\left(x+\frac 1x\right).$ Let $t=x+\frac 1x,\Leftrightarrow t(t^2-3)+10\geq 6t\Leftrightarrow t^3-9t+10=(t-2)(t^2+2t-5)\geq 0.$ Which is obvious by $t\geq 2.\blacksquare$

Hyperabola wrote: Let $a, b>0$. Prove that:$$ (a^3+b^3+a^3b^3)(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} ) +27 \ge 6(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}) $$ The following inequality a bit of stronger. For positives $a$ and $b$ prove that: $$ (a^3+b^3+a^3b^3)(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} ) +45 \ge 9(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}).$$

Let $a, b>0$. Prove that$$ (a^3+b^3+a^3b^3)(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} ) +k\ge \frac{k+9}{6}(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}) $$Where $0\leq k \leq 45.$ $$ \left(a^2+b^2+a^2b^2\right)\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{a^2b^2}\right) + k\ge \frac{k+9}{6}\left(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}\right) $$Where $0\leq k \leq 15.$ $$ \left(a^3+b^3+a^3b^3\right) \left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} \right) +k\geq \frac{k+9}{6}\left(a^2+b^2+\frac{1}{a^2} +\frac{1}{b^2} +\frac{a^2}{b^2} +\frac{b^2}{a^2}\right) $$Where $0\leq k \leq\frac{9}{2}.$

sqing wrote: Let $a, b>0$. Prove that$$ (a^3+b^3+a^3b^3)(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} ) +k\ge \frac{k+9}{6}(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}) $$Where $0\leq k \leq 45.$ Hyperabola wrote: Let $a, b>0$. Prove that:$$ (a^3+b^3+a^3b^3)(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{a^3b^3} ) +27 \ge 6(a+b+\frac{1}{a} +\frac{1}{b} +\frac{a}{b} +\frac{b}{a}) $$ By AM-GM $$x^3 +\frac 1{x^3}=\left(x+\frac 1x\right)^3+8+8-3\left(x+\frac 1x\right)-16\geq 9\left(x+\frac 1x\right)-16$$

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