<s>[INST] Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{9a^2+b^2}}+\frac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \frac32.$ 
Prove that $$\dfrac{a}{\sqrt{a^2+b^2}}+\dfrac{b}{\sqrt{9a^2+b^2}}+\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \dfrac{3}{2}.$$
When is equality attained ?

My Attempt :
I could not think of anything suitable for the entire LHS. The last term $\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2}} $ suggested C-S
. So I applied C-S on the values $\left(\dfrac{2a}{\sqrt{a^2+b^2}}\right)$ and $\left(\dfrac{b}{\sqrt{9a^2+b^2}}\right)$ to get :
$$\left(\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2}}\right)^2\leq \left(\dfrac{4a^2}{a^2+b^2}\right)\left(\dfrac{b^2}{9a^2+b^2}\right)\leq\left(\dfrac{4a^2}{b^2}\right)\times \left(\dfrac{b^2}{9a^2}\right)=\dfrac{4}{9}.\,\,\,\,(♦)$$
For the first term $\dfrac{a}{\sqrt{a^2+b^2}}$, I applied C-S on the terms $\left(a\right)$ and $\left(\dfrac{1}{\sqrt{a^2+b^2}}\right)$ to get :
$$\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)\leq \left(\dfrac{a^2}{a^2+b^2}\right)^{1/2}\leq 1.\,\,\,\,(♣)$$
Using the same logic for the second term, I get :
$$\dfrac{b}{\sqrt{9a^2+b^2}}\leq 1 \,\,\,\,(♠)$$
Adding all the inequalities, I get a very "weak" inequality when compared to the problem.
What is the best way to prove this inequality ?
 [/INST] setting $$x=\sqrt{\frac{a^2}{a^2+b^2}}$$ and $$y=\sqrt{\frac{b^2}{9a^2+b^2}}$$
we get further $$1+\left(\frac{b}{a}\right)^2=\frac{1}{x^2}$$ and $$9\left(\frac{a}{b}\right)^2+1=\frac{1}{y^2}$$ we can eliminate $$\frac{a}{b}$$ and we get $$y^2=\frac{1-x^2}{8x^2+1}$$ thus our inequality is equivalent to
$$x+\sqrt{\frac{1-x^2}{8x^2+1}}+2x\sqrt{\frac{1-x^2}{8x^2+1}}\le \frac{3}{2}$$
this is equivalent to
$$\sqrt{\frac{1-x^2}{8x^2+1}}(1+2x)\le \frac{3}{2}-x$$
squarint and factorizing we obtain $$1/4\,{\frac { \left( 12\,{x}^{2}-8\,x+5 \right)  \left( 2\,x-1
 \right) ^{2}}{8\,{x}^{2}+1}}
\geq 0$$ which is true.
We also assume that $$a,b$$ are positive.
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