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Inequality 39(Christos Patilas)


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. 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.

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