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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
\(\frac{a+b}{ab}\)≥\(\frac{4}{a+b}\) <=> \(\frac{1}{a}\)+\(\frac{1}{b}\)≥\(\frac{4}{a+b}\)
\(\frac{a}{b+2c}+\frac{a}{b+2a}\ge\frac{4a}{2a+2b+2c}=\frac{2a}{a+b+c}\)
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. \(\frac{\left(a+b+c\right)^2}{ab+2ac+bc+2ab+ac+2bc}\ge\frac{\left(a+b+c\right)^2}{3\left(ab+ac+bc\right)}\ge\frac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)