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For $x\in\left(0, \frac{\pi}{2}\right)$ find a minimal value, which the expression

$$\sec x+\csc x+\sec^{2}x+\csc^{2}x$$

can take.

My attempt:

I followed the trigonometrical approach and obtained

$$\sec x+\csc x+\sec^{2}x+\csc^{2}x=\sqrt{\left(2\csc 2x+1\right)^2-1}+4\csc^{2}2x\geq \sqrt{(2+1)^2-1}+4(1)=4+2\sqrt{2}$$

Above was obtained after lot of manipulations with the trigonometrical identities so I am looking for an easy approach to this problem.

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Let $\sin{x}=a$ and $\cos{x}=b$.

Hence, $a^2+b^2=1$ and by AM-GM we obtain: $$\sec x+\csc x+\sec^{2}x+\csc^{2}x=$$ $$=\frac{a+b}{ab}+\frac{1}{a^2b^2}\geq\frac{2\sqrt2}{\sqrt{a^2+b^2}}+\frac{4}{(a^2+b^2)^2}=4+2\sqrt2.$$ The equality occurs for $a=b=\frac{1}{\sqrt2}$, which says that we got a minimal value.

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  • $\begingroup$ To clarify for other folks who might have the same trouble as me: $(a+b) \ge 2 \sqrt{ab}$ and $a^2 + b^2 \ge 2ab \implies \sqrt{a^2 + b^2} \ge \sqrt{2} \sqrt{ab}$. Multiplying gives $(a+b)\sqrt{a^2 + b^2} \ge 2ab\sqrt{2}$. Now divide by $ab\sqrt{a^2 + b^2}$ to get the first terms of the inequality. The second terms can be compared by using $a^2 + b^2 \ge 2ab$ and squaring. $\endgroup$
    – Aryabhata
    Aug 5 '17 at 8:34
  • $\begingroup$ @Aryabhata. It is simply $\sqrt{\frac{a^2+b^2}{2}}\geq \frac{a+b}{2}\geq \sqrt{ab}\geq \frac{2ab}{a+b}$. In particular$\sqrt{\frac{a^2+b^2}{2}}\geq \frac{2ab}{a+b}$. $\endgroup$
    – Maverick
    Aug 5 '17 at 15:43
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    $\begingroup$ @PankajSinha: I was only trying to use AM-GM as noted in the answer. The leftmost part of your inequality is more easily seen using Cauchy-Schwarz. In any case, looks like it is just us two who have noticed this nice answer. $\endgroup$
    – Aryabhata
    Aug 5 '17 at 17:26
  • $\begingroup$ You are right. The answer is really beautiful $\endgroup$
    – Maverick
    Aug 5 '17 at 17:34

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