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# Inequality 39(Christos Patilas)

Problem:

If it holds that $\displaystyle \left(\sin x+\cos x\right)^{k}\leq 8(\sin^{n}x+\cos^{n}x)$, then find the value of

$\displaystyle (n+k)_{\max}$.

1st solution:

From the Power-Mean Inequality we have that

$\displaystyle \sqrt[n]{\frac{\sin^nx+\cos^nx}{2}}\geq \sqrt{\frac{\sin^2x+\cos^2x}{2}}\Longrightarrow \sin^nx+\cos^nx\geq 2\cdot \left(\frac{\sqrt{2}}{2}\right)^{n}$.

So, multiplying by $\displaystyle 8$ we have that

$\displaystyle 8\left(\sin^nx+\cos^nx\right)\geq 16\cdot \left(\frac{\sqrt{2}}{2}\right)^{n}$.

Taking now in hand the left hand side, we know that

$\displaystyle \sin x+\cos x\leq \sqrt{2}\Longrightarrow \left(\sin x+\cos x\right)^{k}\leq \left(\sqrt{2}\right)^{k}$.

So, we know that $\displaystyle 16\cdot \left(\frac{\sqrt{2}}{2}\right)^{n}\geq \left(\sqrt{2}\right)^{k}$. Doing the manipulations on both sides we get that

$\displaystyle n+k\leq 8$, hence $\displaystyle (n+k)_{\max}=8$, Q.E.D.

2nd solution (An idea by Vo Quoc Ba Can):

The inequality is symmetric on $\displaystyle \sin x, \cos x$. So, we only need to find the maximum of those two constants for the values of which

$\displaystyle \sin x=\cos x$, that is $\displaystyle x=\frac{\pi}{4}$. S

o, plugging on the above inequality the value

$\displaystyle x=\frac{\pi}{4}$ we get the desired maximum result, Q.E.D.

## 4 thoughts on “Inequality 39(Christos Patilas)”

1. Smitha says:

You have a great website! 😀 I’ll definitely be visiting again!

• Thank you Smitha for your nice words. I’m trying my best for the inequalities!

2. Silouan says:

I think you need something more in the above solution. Ok you proved that $\displaystyle n+k\leq 8$, but now you have to prove that in case $\displaystyle n+k=8$ the inequality holds.

• Yes silouan you are right. But can you please give me a hint of how to prove this inequality for $\displaystyle n+k=8$ because i can’t think of something to show that. Thanks in advance!