# 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 ?

• As for the equality, it is attained at $b=a\cdot\sqrt3$. – Ivan Neretin Feb 21 '17 at 12:27

## 3 Answers

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.

It's enough to prove our inequality for positives $a$ and $b$.

Let $a^2=\frac{x}{3}$, $b^2=y$ and $x^2+y^2=2kxy$.

Hence, $k\geq1$ and we need to prove that $$\sqrt{\frac{x}{x+3y}}+\sqrt{\frac{y}{3x+y}}+2\sqrt{\frac{xy}{(3x+y)(x+3y)}}\leq\frac{3}{2}$$ or $$\frac{\sqrt{x(3x+y)}+\sqrt{y(x+3y)}}{\sqrt{(3x+y)(x+3y)}}\leq\frac{3}{2}-2\sqrt{\frac{xy}{(3x+y)(x+3y)}}$$ and since $$2\sqrt{\frac{xy}{(3x+y)(x+3y)}}=2\sqrt{\frac{xy}{3(x^2+y^2)+10xy}}\leq2\sqrt{\frac{xy}{6xy+10xy}}=\frac{1}{2}<\frac{3}{2},$$ we need to prove that $$\frac{3x^2+3y^2+2xy+2\sqrt{xy(3x+y)(x+3y)}}{3(x^2+y^2)+10xy}\leq\left(\frac{3}{2}-2\sqrt{\frac{xy}{(3x+y)(x+3y)}}\right)^2$$ or $$\frac{6k+2}{6k+10}+\frac{2}{\sqrt{6k+10}}\leq\left(\frac{3}{2}-\frac{2}{\sqrt{6k+10}}\right)^2$$ or $$15k+49\geq16\sqrt{6k+10},$$ which is AM-GM: $$15k+49=12k+20+3k+29\geq12k+20+32\geq2\sqrt{(12k+20)\cdot32}=16\sqrt{6k+10}$$

Done!

Hint: Apply the Cauchy-Schwarz inequality to the vectors $(2a, \sqrt{a^2+b^2}, 2a)$ and $(\sqrt{9a^2+b^2}, 2b, 2b)$.