Can You Prove This Challenging Inequality for x Between 1.5 and 5?

[anemone]

Here is this week's POTW:

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Suppose $$\frac{3}{2}\le x \le 5.$$ Prove that $$2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}$$.

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[anemone]

Suppose $$\frac{3}{2}\le x \le 5.$$ Prove that $$2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}$$.Congratulations to MarkFL for his correct solution:), which you can find below:

Spoiler

Let:

$$f(x)=2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}$$

Hence:

$$f'(x)=\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{2x-3}}-\frac{3}{2\sqrt{15-3x}}$$

Equating the derivative to zero, and using a numeric root-finding technique, we obtain the critical value:

$$x\approx4.04879336468766$$

Now, we find:

$$f'(4)>0$$ and $$f'(4.1)<0$$

Next, we check the end-points of the domain:

$$f(1.5)\approx6.40265$$

$$f(5)\approx7.54473$$

Thus, by the first derivative test, we conclude:

$$f_{\max}\approx f(4.04879336468766)\approx8.440953705913998489<2\sqrt{19}\approx8.717797887081348$$

Alternate solution:

Spoiler

By the Cauchy–Schwarz inequality, we have

Suppose $$\frac{3}{2}\le x \le 5.$$ Prove that $$2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}$$.

$$\begin{align*}2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}&=\sqrt{x+1}+\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}\\&\le \sqrt{1^2+1^2+1+1^2}\sqrt{\sqrt{x+1}^2+\sqrt{x+1}^2+\sqrt{2x-3}^2+\sqrt{15-3x}^2}\\&\le 2\sqrt{x+14}\\&\le 2\sqrt{19}\end{align*}$$

and equality holds if and only if $$\sqrt{x+1}=\sqrt{2x-3}=\sqrt{15-3x}$$ at $x=5$ but that is impossible.

Thus, we have proved that $$2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}$$.