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[Princess3107]

Cho a,b,c>0.CMR: $ \frac{(b+c-a)^2}{b^2+c^2+bc} + \frac{(c+a-b)^2}{c^2+a^2+ca} + \frac{(a+b-c)^2}{a^2+b^2+ab} \ge 1$

[hanguyen445]

Cho a,b,c>0.CMR: $ \frac{(b+c-a)^2}{b^2+c^2+bc} + \frac{(c+a-b)^2}{c^2+a^2+ca} + \frac{(a+b-c)^2}{a^2+b^2+ab} \ge 1$

Đặt $P=\sum\frac{(b+c-a)^2}{b^2+c^2+bc}$

 

Sử dụng BĐT cauchy-schwarz , ta có đánh giá:

$P+1=P+\frac{a+b+c}{a+b+c}=\sum (\frac{(b+c-a)^2}{b^2+c^2+bc}+\frac{a}{a+b+c})=\sum (\frac{(b+c-a)^2}{b^2+c^2+bc}+\frac{a^2}{a^2+ab+ac})\ge\sum\frac{(b+c)^2}{a^2+b^2+c^2+ab+bc+ac})=Q\\$

Hay $P+1\ge\ Q$ với \[Q = \frac{{{{\left( {b + c} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + bc + ac}} + \frac{{{{\left( {a + b} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + bc + ac}} + \frac{{{{\left( {a + c} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + bc + ac}} = 2\]

Do đó ta có $P+1\ge 2\Leftrightarrow P\ge 1$. Hoàn tất chứng minh.

Edited by hanguyen445, 24-11-2023 - 16:56.