Cho a,b,c là các số thực dương thỏa mãn [tex]c\geq a[/tex]. Cmr :
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[tex]P=(\frac{a}{a+b})^2+(\frac{b}{b+c})^2+4.(\frac{c}{c+a})^2\\\\ =(\frac{1}{1+\frac{b}{a}})^2+(\frac{1}{1+\frac{c}{b}})^2+4.(\frac{1}{1+\frac{c}{a}})^2\\\\ +, (\frac{b}{a};\frac{c}{b})=(x;y)=> xy\geq 1\\\\ => P=\frac{1}{(x+1)^2}+\frac{1}{(y+1)^2}+4.\frac{1}{(1+\frac{1}{xy})^2}\\\\ \geq \frac{1}{2.(x^2+1)}+\frac{1}{2.(y^2+1)}+4.\frac{(xy)^2}{(xy+1)^2}\\\\ \geq \frac{1}{2}.\frac{2}{xy+1}+4.\frac{(xy)^2}{(xy+1)^2}=L\\\\ xy=t (t\geq 1)\\\\ => L=\frac{1}{t+1}+4.\frac{t^2}{(t+1)^2}\geq \frac{3}{2}\\\\ <=> \frac{t+1+4t^2}{(t+1)^2}\geq \frac{3}{2}\\\\ <=> 8t^2+2t+2\geq 3.(t^2+2t+1)\\\\ <=> 8t^2+2t+2\geq 3t^2+6t+3\\\\ <=> 5t^2-4t-1\geq 0\\\\ <=> (t-1).(5t+1)\geq 0 (luôn đúng)[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
