Chủ đề này có 2 trả lời

[mayxanh]

CMR với x,y,x>0

$(1+2\frac{a}{b})^{2}+(1+2\frac{b}{c})^{2}+(1+2\frac{c}{a})^{2} \geq $\frac{9(a+b+c)^2}{ab+bc+ca}$

[quoccuonglqd]

$VT=4[(\frac{a^{2}}{b^{2}}+1)+(\frac{b^{2}}{c^{2}}+1)+(\frac{c^{2}}{a^{2}}+1)]+4(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})-3\geq 12(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})-\frac{(a+b+c)^{2}}{ab+bc+ca}$

Áp dụng C-S:$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{(a+b+c)^{2}}{ab+bc+ca}$

Nên $VT\geq \frac{12(a+b+c)^{2}}{ab+bc+ca}-\frac{3(a+b+c)^{2}}{ab+bc+ca}=\frac{9(a+b+c)^{2}}{ab+bc+ca}$

[takarin1512]

Trước hết, ta luôn có $\sum a^2\geq \sum ab$$\Rightarrow \left ( \sum a \right )^2\geq 3\sum ab\Rightarrow \frac{\left (\sum a \right )^2}{\sum ab}\geq 3$

$\sum \left ( 1+2\frac{a}{b} \right )^2=3+4\sum \frac{a}{b}+4\sum \left ( \frac{a}{b} \right )^2=4\sum \frac{a}{b}+4\sum \left ( \left ( \frac{a}{b} \right )^2+1 \right )-9\geq 12\sum \frac{a}{b}-9=12\sum \frac{a^2}{ab}-9\geq 12\frac{\left ( \sum a \right )^2}{\sum ab}-9\geq 9\frac{\left ( \sum a \right )^2}{\sum ab}$