Chủ đề này có 2 trả lời

[asapdocky]

1. Cho a,b,c > 0 a+b+c = 1. CMR: $\frac{1}{a^{2}+b^{2}+c^{2}} + \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} \geq 30$

2. Cho tam giác ABC. CMR: $\frac{1}{2+cos2A} + \frac{1}{2+cos2B} + \frac{1}{2-cos2C} \geq \frac{6}{5}$

[quan1234]

1. Cho a,b,c > 0 a+b+c = 1. CMR: $\frac{1}{a^{2}+b^{2}+c^{2}} + \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} \geq 30$

2. Cho tam giác ABC. CMR: $\frac{1}{2+cos2A} + \frac{1}{2+cos2B} + \frac{1}{2-cos2C} \geq \frac{6}{5}$

Bài 1 Ta có$a^2+b^2+c^2\geq ab+bc+ac \Rightarrow (a+b+c)^2\geq 3ab+3bc+3ac\Leftrightarrow \frac{(a+b+c)^2}{3}\geq ab+bc+ac$

$\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=(\frac{1}{a^2+b^2+c^2}+\frac{1}{3ab}+\frac{1}{3bc}+\frac{1}{3ac})+(\frac{2}{3ab}+\frac{2}{3bc}+\frac{2}{3ac})\geq \frac{4}{\sqrt[4]{(a^2+b^2+c^2)3ab3ac3bc}}+\frac{6}{\sqrt[3]{3ab3bc3ac}}\geq \frac{16}{a^2+b^2+c^2+3ab+3bc+3ac}+\frac{6}{ab+bc+ac}=\frac{16}{(a+b+c)^2+ab+bc+ac}+\frac{6}{ab+bc+ac}\geq 12+18=30$

 

[GeminiKid]

1. Cho a,b,c > 0 a+b+c = 1. CMR: $\frac{1}{a^{2}+b^{2}+c^{2}} + \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} \geq 30$

2. Cho tam giác ABC. CMR: $\frac{1}{2+cos2A} + \frac{1}{2+cos2B} + \frac{1}{2-cos2C} \geq \frac{6}{5}$

$Bài 1 ta cần 2 bổ đề \frac{1}{x}+\frac{1}{y}\geq \frac{4}{ x+y } và \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{ x+y+z } (dễ cm) Áp dụng ta có \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq \frac{9}{ab+bc+ca} và \frac{1}{a^{2}b^{2}+c^{2}}+\frac{1}{ab+bc+ca}+\frac{1}{ab+ac+ca}\geq \frac{9}{\left ( a+b+c \right )^{2}}= 9(1) Mặt khác ab+bc+ca\leq \frac{\left ( a+b+c \right )^{2}}{3}= \frac{1}{3}\Rightarrow \frac{7}{ab+bc+ca}\geq 21(2)Từ (1) và (2)\Rightarrow VT\geq 30$