Let $a> 0, b> 0$. Prove that: $\frac{{{a}^{2}}}{b}+\frac{{{b}^{2}}}{a}+7\left( a+b \right)\ge 8\sqrt{2\left( {{a}^{2}}+{{b}^{2}} \right)}$

Question

Let $a> 0, b> 0$. Prove that: $$\frac{{{a}^{2}}}{b}+\frac{{{b}^{2}}}{a}+7\left( a+b \right)\ge 8\sqrt{2\left( {{a}^{2}}+{{b}^{2}} \right)}$$ My answer is:

Please give other interesting answers.

Answer 1

after squaring (all is positive) and rearranging and factorizing we get $${\frac { \left( {a}^{2}+18\,ab+{b}^{2} \right) \left( a-b \right) ^{4 }}{{a}^{2}{b}^{2}}} \geq 0$$ which is true.

Answer 2

Let $a^2+b^2=2tab$.

Hence, we need to prove that $$\frac{\sqrt{a^2+b^2+2ab}(a^2+b^2-ab)}{ab}+7\sqrt{a^2+b^2+2ab}\geq8\sqrt{2(a^2+b^2)}$$ or $$\sqrt{2(t+1)}(2t-1)+7\sqrt{2(t+1)}\geq16\sqrt{t}$$ or $$(t+3)\sqrt{2(t+1)}\geq8\sqrt{t},$$ which is true by AM-GM: $$(t+3)\sqrt{2(t+1)}\geq4\sqrt[4]{t\cdot1^3}\sqrt{4\sqrt{t}}=8\sqrt{t}.$$ Done!

Answer 3

$$\frac{a^{2}}{b}+ \frac{b^{2}}{a}+ 7\left ( a+ b \right )= \left ( a+ b \right )\left [ \frac{\left ( a+ b \right )^{2}}{ab}+ \frac{16ab}{\left ( a+ b \right )^{2}}- \frac{16ab}{\left ( a+ b \right )^{2}}+ 4\right ]$$ $$\geq \left ( a+ b \right )\left [ 8- \frac{16ab}{\left ( a+ b \right )^{2}}+ 4 \right ]= 4\left ( a+ b \right )\left [ \frac{2\left ( a^{2}+ b^{2} \right )}{\left ( a+ b \right )^{2}}+ 1 \right ]\geq 8\sqrt{2\left ( a^{2}+ b^{2} \right )}$$