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Let $h(x)=(1-\frac1{x})^{\frac x2}$ for $x\ge2$. Claim: $h(x)\gt\frac12$ if $x\gt2$. Proof: Let $g(x)=\ln(1-x)+x$ for $0\le x\lt1$. Since $g'(x)=-\frac1{1-x} + 1 \le 0$, we have $g(x)\ge g(0)=0$. \$...