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 \(\displaystyle \frac{3}{2}\le x \le 5.\) Prove that \(\displaystyle 2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}\).

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[anemone]

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

Spoiler

Let:

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

Hence:

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

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

\(\displaystyle x\approx4.04879336468766\)

Now, we find:

\(\displaystyle f'(4)>0\) and \(\displaystyle f'(4.1)<0\)

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

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

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

Thus, by the first derivative test, we conclude:

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

Alternate solution:

Spoiler

By the Cauchy–Schwarz inequality, we have

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

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

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

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