Suppose 23≤x≤5. Prove that 2x+1+2x−3+15−3x<219.
Solution
By Cauchy's inequality, we have 2x+1+2x−3+15−3x=x+1+x+1+2x−3+15−3x≤[(x+1)+(x+1)+(2x−3)+(15−3x)](12+12+12+12)=2x+14≤219, and the equality holds if and only if x+1=2x−3=15−3x and x=5. But this is impossible. So 2x+1+2x−3+15−3x<219.
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Source: MathNet,
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