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题:求$\underset{\,}{\,}f:\small\mathbb{R}^+\to\mathbb{R}^+\,$使$\,x^2+y^2+axy-f(a)(x+y)^2\ge 0\;\small(\forall x,y\ge 0,\;\forall a>0)$| 解:取$\underset{\,}{\,}x=y=1\,$得$\,2+a-4f(a)\ge 0,\;f(a)\le \frac{1}{2}+\frac{a}{4}.$B1z!n $\underset{\,}{\qquad}$取$\,x=1,\,y=0\,$得$\;f(a)\le 1.\;$据此令$\;\boxed{f(a)=\min\big(1,{\small\frac{1}{2}+\frac{a}{4}}\big)},\,$则 k:! $(1)\quad a\in (0,2]{\small\implies}f(a)=\frac{1}{2}+\frac{a}{4}\small\implies$2Tz $\qquad\qquad x^2+y^2+axy-f(a)(x+y)^2=\frac{1}{4}(x-y)^2(2-a)\ge 0\underset{\,}{,}$wZT/AT $(2)\quad a>2{\small\implies}f(a)=1{\small\implies x^2+y^2+axy-f(a)(x+y)^2}=(a-2)xy\ge 0.$3{2&
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