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\(\frac{a}{b+2c}+\frac{a}{b+2a}\ge\frac{4a}{2a+2b+2c}=\frac{2a}{a+b+c}\)
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 \(a=b=c\)
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)
\(\Rightarrow2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
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
\(\frac{a+b}{ab}\)≥\(\frac{4}{a+b}\) <=> \(\frac{1}{a}\)+\(\frac{1}{b}\)≥\(\frac{4}{a+b}\)
Từ \(abc\le\left(\frac{a+b+c}{3}\right)^3\Rightarrow a+b+c\ge3\)
Ta có BĐT phụ \(\frac{1}{2a+1}\ge-\frac{2}{9}a+\frac{5}{9}\)
\(\Leftrightarrow\frac{4\left(a-1\right)^2}{9\left(2a+1\right)}\ge0\forall0< a< 3\) (đúng)
Tương tự ta cũng có:
\(\frac{1}{2b+1}\ge-\frac{2}{9}b+\frac{5}{9};\frac{1}{2c+1}\ge-\frac{2}{9}c+\frac{5}{9}\)
Cộng theo vế 3BĐT trên ta có:
\(VT\ge-\frac{2}{9}\left(a+b+c\right)+\frac{5}{9}\cdot3=1=VP\)
Khi a=b=c=1
\(\text{Σ}\frac{c}{2a+2b-c}=\text{Σ}\frac{c^2}{2ac+2bc-c^2}\) (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