Help me prove the exponential inequality $2^{\sin^2 x} + 2^{\cos^2 x} \leq 3$

Question

Please help me with this inequality: $$2^{\sin^2 x} + 2^{\cos^2 x} \leq 3$$ I've reduced it to this: $$2^t + 2^{1-t} \leq 3 \,\,\,\, \text{where}\,\,t=\sin^2 x$$ and I also did a proof that is not elegant at all and kind of informal, I think that my brain after seeing this equation for so long as just frozen and won't do anything. Any new ideas would be appreciated greatly.

Answer 1

Rearranging the equation we have that

$$2^t + 2^{1-t} = 3 \implies 2^{2t} - 3\cdot 2^t + 2 = 0$$

which after factoring gets us

$$\left(2^t-1\right)\left(2^t-2\right) = 0$$

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

$$\left(2^t-1\right)\left(2^t-2\right) \leq 0 \iff 1 \leq 2^t \leq 2 \iff 0\leq t \leq 1$$

which immediately gives us our desired result.

Answer 2

Since $\;2^{\sin^2x}\ge1\;$ and $\;2^{\cos^2x}\ge1\;,\;$ then

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

Moreover,

$2^{\sin^2 x}+2^{\cos^2x}=\left[\left(\sqrt{2}\right)^{\sin^2x}-\left(\sqrt{2}\right)^{\cos^2x}\right]^2+2\sqrt{2}\ge2\sqrt{2}\;.$

Answer 3

I think you're right.

Starting from $2^t +2^{1-t}, 0 \leq t \leq 1$,

\begin{align*} 2^t + 2^{1-t} &= 2^{1/2}\left( 2^{t-1/2} + 2^{-(t-1/2)} \right)\\ &= \sqrt 2 \left( e^{(t-1/2)\ln 2} + e^{-(t-1/2)\ln 2} \right)\\ &= 2\sqrt 2 \cosh\left( (t-1/2)\ln 2 \right)\\ &= 2\sqrt 2 \cosh(u), \end{align*}

where $u = (t-1/2)\ln 2$, so $-\frac{\ln 2}2 \leq u \leq \frac{\ln 2}{2}$. The minimum value of $\cosh$ on this interval is $1$ when $u = 0$, and $3$ when $|u| = \frac{\ln 2}{2}$.

Answer 4

Becomes familiar as $$2^{\frac{1}{2} - w} + 2^{\frac{1}{2} + w} = \sqrt 2 \left( 2^{-w} + 2^w \right) = \sqrt 2 \left( \frac{1}{v} + v \right) $$ where we can take $v = 2^w$ with $0 \leq w \leq \frac{1}{2}$ so that $1 \leq v \leq \sqrt 2$ and $$ 2 \leq v + \frac{1}{v} \leq \frac{3}{2} \sqrt 2 $$ and the original quantity is at least $\sqrt 8$ and no larger than $3$

About $v + \frac{1}{v}$ for $v \geq 1,$ take $1 \leq p < q$ so that $pq > 1.$ Then $$ \left(q + \frac{1}{q} \right) - \left(p + \frac{1}{p} \right)= (q-p) - \frac{q-p}{pq} > (q-p) - (q-p) = 0 $$

Answer 5

Maybe the following reasoning is elegant.

It's enough to prove our inequality for $x\in\left[0,\frac{\pi}{4}\right]$ and

since $f(t)=2^t$ is a convex function and $(1,0)\succ(\cos^2x,\sin^2x),$ by Karamata we obtain: $$2^{\cos^2x}+2^{\sin^2x}\leq2^1+2^0=3.$$