hello, after squaring two times we get $ \left( 49\,{b}^{6}+54\,{b}^{5}+155\,{b}^{4}+68\,{b}^{3}+139\,{b}^{2}+ 14\,b+49 \right) \left( b-1 \right) ^{2} \geq0$ which is true. Sonnhard.

Using $\sqrt{xy} \le \frac{x+y}{2}$, we get $LHS \le \frac{1}{2} (2a^2+2ab+(a+b)^2+2b^2+a^2+b^2)$ We just have to prove that $6(a^2+b^2) \ge (3a^2+3b^2+(a+b)^2+2ab)$ which is equivalent to $2a^2+2b^2 \ge 4ab$ which is true.

Son hard, how did you get rid of the variable a? Did you set it as 1 or something?

you can set a equal to 1, since that just scales everything by 1/a^2 (the map from (a,b) to (1, b/a)) since the equation is homogeneous

$\sqrt{2}(\sqrt{a(a+b)^3}+b\sqrt{a^2+b^2})$ $=\sqrt{(2a^2+2ab)(a+b)^2}+\sqrt{2b^2(a^2+b^2)}$ $\leq \sqrt{(3a^2+b^2)(2a^2+2b^2)}+\sqrt{2b^2(a^2+b^2)}$ $\leq \frac{3a^2+b^2+2a^2+2b^2}{2}+ \frac{2b^2+a^2+b^2}{2}= 3(a^2+b^2).$

After squaring we need to show, $7(a^4+b^4) + 10a^2b^2 \geq 6a^3b+2ab^3+4\sqrt{ab^2(a+b)^3(a^2+b^2)}$ We have $ab^2(a+b)^3(a^2+b^2)=(b^4+2ab^3+a^2b^2)(a^4+a^3b+a^2b^2+ab^3)$ So $4\sqrt{ab^2(a+b)^3(a^2+b^2)} \leq 2(b^4+2ab^3+a^2b^2+a^4+a^3b+a^2b^2+ab^3)$ Now it is sufficient to show $5(a^4+b^4) + 6a^2b^2 \geq 8(ab^3+ba^3)$....$(\bigstar)$ $3(a^4+a^2b^2) \geq 6a^3b$ ....$(i)$ $3(b^4+a^2b^2) \geq 6b^3a$....$(ii)$ $2(a^4+b^4) \geq 2(ab^3+ba^3)$....$(iii)$ Adding $(i),(ii),(iii)$ we get $(\bigstar)$. Hence we are done.

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sqing wrote: $\sqrt{2}(\sqrt{a(a+b)^3}+b\sqrt{a^2+b^2})$ $=\sqrt{(2a^2+2ab)(a+b)^2}+\sqrt{2b^2(a^2+b^2)}$ $\leq \sqrt{(3a^2+b^2)(2a^2+2b^2)}+\sqrt{2b^2(a^2+b^2)}$ $\leq \frac{3a^2+b^2+2a^2+2b^2}{2}+ \frac{2b^2+a^2+b^2}{2}= 3(a^2+b^2).$ Awesome solution!