Pamela Meyer

Answered

2022-01-17

Help me prove the exponential inequality

${2}^{{\mathrm{sin}}^{2}x}+{2}^{{\mathrm{cos}}^{2}x}\le 3$

Ive

Ive

Answer & Explanation

Bernard Lacey

Expert

2022-01-16Added 30 answers

Since ${2}^{{\mathrm{sin}}^{2}x}\ge 1$ and ${2}^{{\mathrm{cos}}^{2}x}\ge 1$ , then

${2}^{{\mathrm{sin}}^{2}x}+{2}^{{\mathrm{cos}}^{2}x}=3-({2}^{{\mathrm{sin}}^{2}x}-1)\cdot ({2}^{{\mathrm{cos}}^{2}x}-1)\le 3$

Moreover,

$2}^{{\mathrm{sin}}^{2}x}+{2}^{{\mathrm{cos}}^{2}x}={[{\left(\sqrt{2}\right)}^{{\mathrm{sin}}^{2}x}-{\left(\sqrt{2}\right)}^{{\mathrm{cos}}^{2}x}]}^{2}+2\sqrt{2}\ge 2\sqrt{2$ .

Moreover,

RizerMix

Expert

2022-01-19Added 437 answers

Rearranging the equation we have that

which after factoring gets us

This is a parabola with roots at 1 and 2 that opens upward. Therefore we have that

which immediately gives us our desired result.

alenahelenash

Expert

2022-01-24Added 366 answers

I think youre

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