Inequality: 2^{sin x} + 2^{cos x} — Sharp Lower Bound | AzuCATion

Which statement is true for all real \(x\)?

\[ \text{(a)}\;2^{\sin x}+2^{\cos x}>2^{\,1+\tfrac{1}{\sqrt2}} \quad \text{(b)}\;2^{\sin x}+2^{\cos x}>2^{\,1-\tfrac{1}{\sqrt2}} \] \[ \text{(c)}\;2^{\sin x}+2^{\cos x}\ge 2^{\,1+\tfrac{1}{\sqrt2}} \quad \text{(d)}\;{\,2^{\sin x}+2^{\cos x}\ge 2^{\,1-\tfrac{1}{\sqrt2}}\,} \]

AM ≥ GM Trigonometric bounds Always-true inequality

📘 Core Idea (AM ≥ GM + bound on \(\sin x+\cos x\))

By AM ≥ GM, \[ \frac{2^{\sin x}+2^{\cos x}}{2}\;\ge\;\sqrt{\,2^{\sin x}\cdot 2^{\cos x}\,} \;=\;2^{\frac{\sin x+\cos x}{2}}. \] Hence \[ 2^{\sin x}+2^{\cos x}\;\ge\;2\cdot 2^{\frac{\sin x+\cos x}{2}} =2^{\,1+\frac{\sin x+\cos x}{2}}. \] Now use \(\sin x+\cos x=\sqrt2\sin\!\left(x+\tfrac{\pi}{4}\right)\in[-\sqrt2,\sqrt2]\). The minimum of the exponent \(1+\tfrac{\sin x+\cos x}{2}\) is \(1-\tfrac{1}{\sqrt2}\). Therefore, for all \(x\), \[ \boxed{\,2^{\sin x}+2^{\cos x}\;\ge\;2^{\,1-\tfrac{1}{\sqrt2}}\,}. \]

Max–min tip: In general, to extremize \(\sin x+\cos x\), check \(x=\tfrac{\pi}{4}\) (i.e., \(45^\circ\)) for the maximum \(+\sqrt2\) and \(x=\tfrac{5\pi}{4}\) (i.e., \(225^\circ\)) for the minimum \(-\sqrt2\). This follows from \(\sin x+\cos x=\sqrt2\sin(x+\tfrac{\pi}{4})\).

🧠 Why \(\sin x+\cos x\in[-\sqrt2,\sqrt2]\)

\[ \sin x+\cos x =\sqrt2\!\left(\tfrac{1}{\sqrt2}\sin x+\tfrac{1}{\sqrt2}\cos x\right) =\sqrt2\sin\!\left(x+\tfrac{\pi}{4}\right), \] \[ \Rightarrow\quad -\sqrt2\le\sin x+\cos x\le\sqrt2. \]

Thus the exponent \(1+\frac{\sin x+\cos x}{2}\) ranges from \(1-\tfrac{1}{\sqrt2}\) to \(1+\tfrac{1}{\sqrt2}\). Our inequality uses the least exponent to form a universal lower bound.