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Cho \(a=b=c\)
\(\Rightarrow2\left(\frac{a}{a+2a}+\frac{a}{a+2a}+\frac{a}{a+2a}\right)\ge1+\frac{a}{a+2a}+\frac{a}{a+2a}+\frac{a}{a+2a}\)
\(\Leftrightarrow2\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)\ge1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\)
\(\Leftrightarrow2\ge2\) ( Đúng)
\(\Rightarrow2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
\(\text{Σ}\frac{c}{2a+2b-c}=\text{Σ}\frac{c^2}{2ac+2bc-c^2}\) (1)
Áp dụng BDT Cauchy-Schwarz, ta dc:
\(\left(1\right)\ge\frac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)-a^2-b^2-c^2}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)+a^2+b^2+c^2}=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Dấu = xảy ra <=> a=b=c
a) Dùng (a+b)2≥4ab
Chia hai vế cho a+b ( vì ab khác 0)
Ta có a+b≥\(\frac{4ab}{a+b}\) (Chuyển ab sang a+b) ta có
\(\frac{a+b}{ab}\)≥\(\frac{4}{a+b}\) <=> \(\frac{1}{a}\)+\(\frac{1}{b}\)≥\(\frac{4}{a+b}\)
Cho $a, b, c > 0$. Đặt $P = \frac{a}{a+2b+2c} + \frac{b}{2a+b+2c} + \frac{c}{2a+2b+c}$.
Ta cần chứng minh: $\frac{3}{5} \le P < 1$.
* Chứng minh $P \ge \frac{3}{5}$:
Áp dụng bất đẳng thức Cauchy-Schwarz dạng phân thức:
$P = \frac{a^2}{a(a+2b+2c)} + \frac{b^2}{b(2a+b+2c)} + \frac{c^2}{c(2a+2b+c)}$
$\ge \frac{(a + b + c)^2}{a(a+2b+2c) + b(2a+b+2c) + c(2a+2b+c)}$
Rút gọn mẫu thức:
$a(a+2b+2c) + b(2a+b+2c) + c(2a+2b+c)$
$= a^2 + 2ab + 2ac + 2ab + b^2 + 2bc + 2ac + 2bc + c^2$
$= a^2 + b^2 + c^2 + 4ab + 4bc + 4ca$
$= (a + b + c)^2 + 2(ab + bc + ca)$
Ta có bất đẳng thức quen thuộc: $ab + bc + ca \le \frac{(a + b + c)^2}{3}$
Do đó mẫu thức:
$(a + b + c)^2 + 2(ab + bc + ca) \le (a + b + c)^2 + 2 \cdot \frac{(a + b + c)^2}{3} = \frac{5}{3}(a + b + c)^2$
Suy ra: $P \ge \frac{(a + b + c)^2}{\frac{5}{3}(a + b + c)^2} = \frac{3}{5}$
Dấu "=" xảy ra khi $a = b = c$.
Vì $a, b, c > 0$ nên:
$a + 2b + 2c > a + b + c \Rightarrow \frac{a}{a+2b+2c} < \frac{a}{a+b+c}$
$2a + b + 2c > a + b + c \Rightarrow \frac{b}{2a+b+2c} < \frac{b}{a+b+c}$
$2a + 2b + c > a + b + c \Rightarrow \frac{c}{2a+2b+c} < \frac{c}{a+b+c}$
Cộng từng vế của ba bất đẳng thức trên:
$P < \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c}$
$\Rightarrow P < \frac{a + b + c}{a + b + c} = 1$
Vậy $\frac{3}{5} \le \frac{a}{a+2b+2c} + \frac{b}{2a+b+2c} + \frac{c}{2a+2b+c} < 1$ (điều phải chứng minh).
\(\frac{a}{b+2c}+\frac{a}{b+2a}\ge\frac{4a}{2a+2b+2c}=\frac{2a}{a+b+c}\)
Tương tự: \(\frac{b}{c+2a}+\frac{b}{c+2b}\ge\frac{2b}{a+b+c}\) ; \(\frac{c}{a+2b}+\frac{c}{a+2c}\ge\frac{2c}{a+b+c}\)
Cộng vế với vế:
\(\Rightarrow\frac{1}{2}.VT+\frac{a}{b+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}\ge2\)
\(\Leftrightarrow VT+\frac{2a}{b+2a}+\frac{2b}{c+2b}+\frac{2c}{a+2c}\ge4\)
\(\Leftrightarrow VT+\left(1-\frac{b}{b+2a}\right)+\left(1-\frac{c}{c+2b}\right)+\left(1-\frac{a}{a+2c}\right)\ge4\)
\(\Leftrightarrow VT\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
Dấu "=" xảy ra khi \(a=b=c\)
\(\frac{1}{2a-1}+\frac{1}{1}\ge\frac{4}{2a}=\frac{2}{a}\) ; \(\frac{1}{2b-1}+\frac{1}{1}\ge\frac{2}{b}\) ; \(\frac{1}{2c-1}+\frac{1}{1}\ge\frac{2}{c}\)
\(\Rightarrow VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(\Rightarrow VT\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(VT=\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\)
\(=\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}+\frac{c^2}{ac+2bc}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel:
\(\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}+\frac{c^2}{ac+2bc}\ge\frac{\left(a+b+c\right)^2}{ab+2ac+bc+2ab+ac+2bc}\)\(=\frac{\left(a+b+c\right)^2}{3\left(ab+ac+bc\right)}\)\(=\frac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)
Ấy chết, nhầm. \(\frac{\left(a+b+c\right)^2}{ab+2ac+bc+2ab+ac+2bc}\ge\frac{\left(a+b+c\right)^2}{3\left(ab+ac+bc\right)}\ge\frac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)
Tại sao \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) vậy?
lp 9 chưa học cái này hả ????