Cho các số thực dương x, y, z thỏa mãn: [tex]\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}=2015[/tex]
Tìm GTNN của biểu thức: T=[tex]\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}[/tex]
Mọi người giúp em với ạ. Xin cảm ơn mọi người^^
[tex]y^2+z^2\geq \frac{(y+z)^2}{2} => y+z\leq \sqrt{2}.\sqrt{y^2+z^2}\\\\ => \frac{x^2}{y+z}\geq \frac{x^2}{\sqrt{2}.\sqrt{y^2+z^2}}\\\\ => T\geq \frac{x^2}{\sqrt{2}.\sqrt{y^2+z^2}}+\frac{y^2}{\sqrt{2}.\sqrt{x^2+z^2}}+\frac{z^2}{\sqrt{2}.\sqrt{x^2+y^2}}=P\\\\ (\sqrt{x^2+y^2};\sqrt{y^2+z^2};\sqrt{z^2+x^2})=(a;b;c)\\\\ => a+b+c=2015\\\\ => P=\frac{a^2+c^2-b^2}{2\sqrt{2}.b}+\frac{a^2+b^2-c^2}{2\sqrt{2}.c}+\frac{b^2+c^2-b^2}{2\sqrt{2}.a}\\\\ =\frac{1}{2\sqrt{2}}.(\frac{a^2}{b}+\frac{c^2}{b}+\frac{a^2}{c}+\frac{b^2}{c}+\frac{b^2}{a}+\frac{c^2}{a}-a-b-c)\\\\ \geq \frac{1}{2\sqrt{2}}.[\frac{(a+c+a+b+b+c)^2}{b+b+c+c+a+a}-(a+b+c)]\\\\ => P\geq \frac{1}{2\sqrt{2}}.[2.(a+b+c)-(a+b+c)]=\frac{1}{2\sqrt{2}}.(a+b+c)=\frac{2015}{2\sqrt{2}}\\\\ => T\geq P\geq \frac{2015}{2\sqrt{2}}[/tex]
dấu "=" <=> [tex]a=b=c=... <=> ... <=> x=y=z=\frac{2015}{3\sqrt{2}}[/tex]
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