f(x)=ax2+bx+c are positive, and a+b+c=1Prove f(x1)f(x2)⋯f(xn)≥1 is true for all positive x, satisying x1x2⋯xn=1

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Ashley Olson · Accepted Answer

Step 1It&#039;s wrong.For&amp;nbsp;n=2&amp;nbsp;the following is true.minx&amp;gt;0,y&amp;gt;0,xy=1(5x2−14x+10)(5y2−14y+10)=12.The equality occurs for example, for(x,y)=(21−4120,21+4120).By the way,&amp;nbsp;a≥0&amp;nbsp;and&amp;nbsp;c≥0&amp;nbsp;by the given.Also, for&amp;nbsp;b≥0&amp;nbsp;we can use Holder here:∏k=1n(axk2+bxk+c)≥(a∏k=1nxk2n+b∏k=1nxkn+c)n=(a+b+c)n=1

Jared Kemp · Accepted Answer

Step 1(Compare A Quadratic With Positive Coefficients or Inequalities with polynomials on AoPS). Any polynomial f with real positive coefficients satisfiesf(x1)⋯f(xn)≥f(x1⋯xnn)nfor positive numbers&amp;nbsp;x1,…,xn. That is a consequence of the generalized Holder inequality, alternatively one can prove that&amp;nbsp;y→log⁡(f(ey))&amp;nbsp;is convex and apply Jensen&#039;s inequality.In our case is&amp;nbsp;x1⋯xn=1&amp;nbsp;and&amp;nbsp;x1⋯xn=1, so that&amp;nbsp;x1⋯xn=1.

\(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{{2}}}+{b}{x}+{c}\) are positive, and \(\displaystyle{a}+{b}+{c}={1}\) Prove

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