Cho a,b,c dương thỏa mãn a+b+c=1. Tìm max:
P=$\sqrt{abc}(\frac{1}{a+bc}+\frac{1}{b+ac})-\frac{c}{c+ab}$
Cho a,b,c dương thỏa mãn a+b+c=1. Tìm max:
P=$\sqrt{abc}(\frac{1}{a+bc}+\frac{1}{b+ac})-\frac{c}{c+ab}$
Cho a,b,c dương thỏa mãn a+b+c=1. Tìm max:
P=$\sqrt{abc}(\frac{1}{a+bc}+\frac{1}{b+ac})-\frac{c}{c+ab}$
+)Từ giả thiết ta viết lại như sau
$a+b+c=\sqrt{\frac{ab}{c}}.\sqrt{\frac{ac}{b}}+\sqrt{\frac{bc}{a}}.\sqrt{\frac{ba}{c}}+\sqrt{\frac{ca}{b}}.\sqrt{\frac{cb}{a}}$=1
Do đó đặt $x=\sqrt{\frac{bc}{a}},y=\sqrt{\frac{ac}{b}},z=\sqrt{\frac{ab}{c}}$
Vậy xy+yz+zx=1
+)Lúc đó P=$\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}-\frac{1}{z^{2}+1}$
+)Do xy+yz+zx=1 nên tồn tại tam giác sao cho $x=tan\frac{A}{2},y=tan\frac{B}{2},z=tan\frac{C}{2}$
Lúc đó P=$\frac{tan\frac{A}{2}}{tan^{2}\frac{A}{2}+1}+\frac{tan\frac{B}{2}}{tan^{2}\frac{B}{2}+1}-\frac{1}{tan^{2}\frac{C}{2}+1}$
$\Leftrightarrow P=\frac{1}{2}(sinA+sinB)-cos^{2}\frac{C}{2}=sin(\frac{A+b}{2})cos(\frac{A-b}{2})-cos^{2}\frac{C}{2}\leq cos\frac{C}{2}-cos^{2}\frac{C}{2}\leq \frac{1}{4}$
Dấu '=' xẩy ra khi $c=7-4\sqrt{3},a=b=2\sqrt{3}-3$
$\sqrt{O}$ve math
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