For every $x\in[\frac{3}{2}, 5]$ prove that: $(\sqrt{2x-3}+\sqrt{15-3x}+2\sqrt{x+1})^2<71.25$ For every $x\in[\frac{3}{2}, 5]$ prove that: $(\sqrt{2x-3}+\sqrt{15-3x}+2\sqrt{x+1})^2<71.25$
I proved this by saying that $(\sqrt{2x-3}+\sqrt{15-3x}+2\sqrt{x+1})^2=(\sqrt{2}*\sqrt{x-\frac{3}{2}}+\sqrt{\frac{3}{2}}*\sqrt{10-2x}+2\sqrt{x+1})^2$
$\le [(\sqrt{2})^2+(\sqrt{\frac{3}{2}})^2+2^2][(\sqrt{x-\frac{3}{2}})^2+(\sqrt{10-2x})^2+(\sqrt{x+1})^2$ (B.C.S)
$=71.25$
It took me too long till I used $\sqrt{2}, \sqrt{\frac{3}{2}}, 2$ as the numbers for the second bracket used in BCS. Could you please explain to me why I should have immediately intuitively thought of using them and not any other set of numbers?
 A: We'll choose positives $\alpha$, $\beta$ and $\gamma$ such that after using C-S we'll get:$$\left(\sqrt{2x-3}+\sqrt{15-3x}+2\sqrt{x+1}\right)^2=$$
$$=\left(\sqrt{\alpha}\sqrt{\frac{2x-3}{\alpha}}+\sqrt{\beta}\sqrt{\frac{15-3x}{\beta}}+\sqrt{\gamma}\sqrt{\frac{4x+4}{\gamma}}\right)^2\leq$$
$$\leq(\alpha+\beta+\gamma)\left(\frac{2x-3}{\alpha}+\frac{15-3x}{\beta}+\frac{4x+4}{\gamma}\right)=71.25,$$
For which we need: $$\frac{2}{\alpha}-\frac{3}{\beta}+\frac{4}{\gamma}=0$$ and
$$(\alpha+\beta+\gamma)\left(\frac{-3}{\alpha}+\frac{15}{\beta}+\frac{4}{\gamma}\right)=\frac{285}{4}.$$
The first gives $$\beta=\frac{3\alpha\gamma}{2(2\alpha+\gamma)}$$ and we obtain:
$$\left(\alpha+\frac{3\alpha\gamma}{2(2\alpha+\gamma)}+\gamma\right)\left(\frac{-3}{\alpha}+\frac{15}{\frac{3\alpha\gamma}{2(2\alpha+\gamma)}}+\frac{4}{\gamma}\right)=\frac{285}{4}$$ or
$$(2\alpha-\gamma)(96\alpha^2+7\alpha\gamma-28\gamma^2)=0$$ and we can choose $$\gamma=2\alpha,$$ which gives $$\beta=\frac{3}{4}\alpha$$ and for $\alpha=2$ we obtain your solution.
Easy to see that $71.25$ is not a maximal value.
Indeed, the equality should be occur for
$$(\sqrt{\alpha},\sqrt{\beta},\sqrt{\gamma})||\left(\sqrt{\frac{2x-3}{\alpha}},\sqrt{\frac{15-3x}{\beta}},\sqrt{\frac{4x+4}{\gamma}}\right),$$
which gives $$\frac{\sqrt{2x-3}}{\alpha}=\frac{\sqrt{15-3x}}{\beta}=\frac{\sqrt{4x+4}}{\gamma}$$ or
$$\frac{\sqrt{2x-3}}{\alpha}=\frac{\sqrt{15-3x}}{\frac{3}{4}\alpha}=\frac{\sqrt{4x+4}}{2\alpha}.$$
From the first and the third fractions we obtain $x=4$, which not so plays with the second fraction.
