不等式証明(1)

求証﹕\(\log_2(2^x+2^y)\geq\frac{x+y}{2}+1\).

証明﹕由AM-GM不等式得\(2^x+2^y\geq 2\sqrt{2^x\cdot 2^y}=2\sqrt{2^{x+y}}\)

\(\because y=\log_2x\)是增函數

\(\therefore \log_2(2^x+2^y)\geq \log_2{2\sqrt{2^{x+y}}}=\log_2 2+\log_2 2^{\frac{x+y}{2}}=1+\frac{x+y}{2}\)

証畢