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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(B=\frac{1}{(a+2b)(a+2c)}+\frac{1}{(b+2a)(b+2c)}+\frac{1}{(c+2a)(c+2b)}\)
\(\geq \frac{9}{(a+2b)(a+2c)+(b+2a)(b+2c)+(c+2a)(c+2b)}\)
\(\Leftrightarrow B\geq \frac{9}{(a^2+2ac+2ab+4bc)+(b^2+2bc+2ab+4ac)+(c^2+2bc+2ac+4ab)}\)
\(\Leftrightarrow B\geq \frac{9}{a^2+b^2+c^2+8(ab+bc+ac)}=\frac{9}{(a+b+c)^2+6(ab+bc+ac)}(*)\)
Theo hệ quả quen thuộc của BĐT Cô-si:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow (a+b+c)^2\geq 3(ab+bc+ac)\)
\(\Rightarrow 2(a+b+c)^2\geq 6(ab+bc+ac)(**)\)
Từ \((*); (**)\Rightarrow B\geq \frac{9}{(a+b+c)^2+2(a+b+c)^2}=\frac{3}{(a+b+c)^2}\geq \frac{3}{3^2}=\frac{1}{3}\)
(do \(a+b+c\leq 3)\)
Do đó: \(B_{\min}=\frac{1}{3}\)
Dấu bằng xảy ra khi \(a=b=c=1\)
áp dụng BĐT cô si dạng engel cho 3 số dương, ta có:
\(\dfrac{\left(2b+3c\right)^2}{a}+\dfrac{\left(2c+3a\right)^2}{b}+\dfrac{\left(2a+3b\right)^2}{c}\ge\dfrac{\left(5a+5b+5c\right)^2}{a+b+c}=\dfrac{25\left(a+b+c\right)^2}{a+b+c}=25\left(a+b+c\right)\left(đpcm\right)\)
Áp dụng bất đẳng thức Svacxo ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\)
Tương tự : \(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c};\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{a+2b}+\dfrac{3}{b+2c}+\dfrac{3}{c+2a}\)
Dấu = xảy ra khi a=b=c
\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\)(BĐT Cauchy Schawarz)(1)
tương tự \(=>\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c}\left(2\right)\)
\(=>\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\left(3\right)\)
(1)(2)(3)
\(=>3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)
\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\left(dpcm\right)\)
\(P=\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\)
Áp dụng BĐT Cô-si vào 3 số dương ta có :
\(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b}{3}+\dfrac{2c+a}{9}\ge3\sqrt[3]{\dfrac{a^3}{b\left(2c+a\right)}.\dfrac{b}{3}.\dfrac{2c+a}{9}}=a\) ( 1 )
Tương tự ta có :
\(\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c}{3}+\dfrac{2a+b}{9}\ge3\sqrt[3]{\dfrac{b^3}{c\left(2a+b\right)}.\dfrac{c}{3}.\dfrac{2a+b}{9}}=b\) ( 2 )
\(\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{a}{3}+\dfrac{2b+c}{9}\ge3\sqrt[3]{\dfrac{c^3}{a\left(2b+c\right)}.\dfrac{a}{3}.\dfrac{2b+c}{9}}=c\) ( 3 )
Cộng từng vế của ( 1 ) ( 2 ) và ( 3 ) ta có :
\(\dfrac{a^3}{c\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{2}{3}\left(a+b+c\right)\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{2}{3}.3\ge3\)
\(\Leftrightarrow P\ge1\)
\(\LeftrightarrowĐpcm.\)
Dấu " = " xảy ra khi \(a=b=c=1\)
Chúc bạn học tốt
có a3 kìa sao ko thay vào thành aa+b+c r` giải thử nhỉ :D
\(\dfrac{a^3}{\left(a+2b\right)\left(b+2c\right)}+\dfrac{a+2b}{27}+\dfrac{b+2c}{27}\ge3\sqrt[3]{\dfrac{a^3\left(a+2b\right)\left(b+2c\right)}{27^2.\left(a+2b\right)\left(b+2c\right)}}=\dfrac{a}{3}\)
Tương tự:
\(\dfrac{b^3}{\left(b+2c\right)\left(c+2a\right)}+\dfrac{b+2c}{27}+\dfrac{c+2a}{27}\ge\dfrac{b}{3}\)
\(\dfrac{c^3}{\left(c+2a\right)\left(a+2b\right)}+\dfrac{c+2a}{27}+\dfrac{a+2b}{27}\ge\dfrac{c}{3}\)
Cộng vế:
\(VT+\dfrac{2\left(a+b+c\right)}{9}\ge\dfrac{a+b+c}{3}\)
\(\Rightarrow VT\ge\dfrac{a+b+c}{9}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)