The inequality is equivalent to $$\sqrt{\frac{4\cdot\frac{1}{4}+5\cdot a^2}{9}}+5\sqrt{\frac{8\cdot\frac{1}{4}+b^2}{9}}\geq\frac{4\cdot\frac{1}{2}+5a}{9}+5\cdot\frac{8\cdot\frac{1}{2}+b}{9}$$WLOG: $a, b\geq 0$ and the inequality follows from the $QM-AM$ inequality.

now it is corect (nice and easy)

Can someone do this with Cauchy-Schwartz inequality?

involve_solve wrote: Can someone do this with Cauchy-Schwartz inequality? $(1+5a^2)(1+\dfrac{5}{4}) \geq (1+\dfrac{5}{2}a)^2$ and $(2+b^2)(2+\dfrac{1}{4}) \geq (2+\dfrac{1}{2}b)^2$

microsoft_office_word wrote: Let $ a,b\in\mathbb{R}$ such that $ a+b=1 $ . Prove that $ \sqrt{1+5a^2}+5\sqrt{2+b^2} \ge9 $ . Proof of Zhangyunhua:

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microsoft_office_word wrote: Let $ a,b\in\mathbb{R}$ such that $ a+b=1 $ . Prove that $ \sqrt{1+5a^2}+5\sqrt{2+b^2} \ge9 $ . After squaring we need to prove that $$a^2+5b^2+2\sqrt{(1+5a^2)(2+b^2)}\geq6,$$which is true by C-S: $$a^2+5b^2+2\sqrt{(1+5a^2)(2+b^2)}=a^2+5b^2+2\sqrt{(1+4a^2+a^2)(1+1+b^2)}\geq a^2+5b^2+2(1+2a+ab)=$$$$=a^2+5b^2+2(a+b)^2+4a(a+b)+2ab=7a^2+7b^2+10ab=6(a+b)^2+(a-b)^2\geq6.$$

By Cauchy squartz or by $AM\ge HM$ We get. To prove $\sqrt{1+5a^2}+5\sqrt{2+b^2}\ge 9$ Let $P(a,b,c)$ be lhs $P(a,b,c)\ge \sqrt{( 1+5a^2+ 10+5b^2)(6)}$ $6=1^2+1^2+1^2+1^2+1^2+1^2$ But we know $2(a^2+b^2)\ge (a+b)^2$ $\implies a^2+b^2\ge \frac{1}{2}$ Then put it in our original eqn $P(a,b,c)\ge \sqrt{(11+\frac{5}{2})(6)}$ $\implies P(a,b,c)\ge 3^2=9$