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let $a+b+c=1,a,b,c\ge 0$,find this following maximum $$f(a,b,c)=Aa^2+Bb^2+Cc^2$$ where $A,B,C$ be postive constant numbers.

My idea: if find this minimum value,I can find it,because we have use Cauchy-Schwarz inequality,then we have $$(Aa^2+Bb^2+Cc^2)(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C})\ge (a+b+c)^2=1$$ so $$Aa^2+Bb^2+Cc^2\ge\dfrac{ABC}{AB+BC+AC}$$ But find this maximum,I can't to solve it ,Thank you for you help,

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  • $\begingroup$ "find this maximum, I can't" You sound like yoda :) $\endgroup$ – Guy Mar 14 '14 at 11:12
  • $\begingroup$ Hint: Use $AM(x,y,z)\le \text{max}\{x,y,z\}$. $\endgroup$ – walcher Mar 14 '14 at 11:18
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    $\begingroup$ Are you allowed to use Lagrange multipliers ? $\endgroup$ – Claude Leibovici Mar 14 '14 at 11:18
  • $\begingroup$ @ClaudeLeibovici,I don't think use Lagrange multipliers to solve this form inequality $\endgroup$ – user94270 Mar 14 '14 at 11:22
  • $\begingroup$ Set $x^2=Aa^2, y^2=Bb^2$ and $z^2=Cc^2$, then the problem reduces to maximizing $x^2+y^2+z^2$ on the set of points $\frac{x}{\sqrt A}+\frac{y}{\sqrt B}+\frac{z}{\sqrt C}=1$, and $x,y,z>0$. Thats maximizing distance between origin and the part of the plane $$\frac{x}{\sqrt A}+\frac{y}{\sqrt B}+\frac{z}{\sqrt C}=1$$ in the first quadrant. Should occur at one of the vertices of the triangle :) $\endgroup$ – r9m Mar 14 '14 at 11:23
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Let $M=\text{max}\{A.B.C\}$, then $$ Aa^2+Bb^2+Cc^2\le M(a^2+b^2+c^2)\le M(a+b+c)^2=M $$ Obviously $M$ is attained, because $A,B$ and $C$ are attained, hence $M$ is the maximum value of your function.

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