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\(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\Rightarrow b=\frac{2ac}{a+c}\)
ta có: \(P=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}=\frac{\frac{a^2+3ac}{a+c}}{\frac{2a^2}{a+c}}+\frac{\frac{c^2+3ac}{a+c}}{\frac{2c^2}{a+c}}\)
\(=\frac{a^2+3ac}{2a^2}+\frac{c^2+3ac}{2c^2}=1+\frac{3}{2}\left(\frac{c}{a}+\frac{a}{c}\right)\ge1+\frac{3}{2}\cdot2\sqrt{\frac{c}{a}\cdot\frac{a}{c}}=4\)
Dấu "=" xảy ra khi a=b=c
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq \left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)^2\)
\(\Rightarrow \frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ac}{abc}=ab+bc+ac\)
Do đó:
\(P\geq ab+bc+ac+\frac{9}{2(a+b+c)}\)
Áp dụng BĐT AM-GM:
\(ab+bc+ac+\frac{9}{2(a+b+c)}=\frac{ab+bc+ac}{2}+\frac{ab+bc+ac}{2}+\frac{9}{2(a+b+c)}\geq 3\sqrt[3]{\frac{9(ab+bc+ac)^2}{8(a+b+c)}}\)
Theo một kết quả quen thuộc của BĐT AM-GM:
\((ab+bc+ac)^2\geq 3abc(a+b+c)\)
Thay \(abc=1\Rightarrow (ab+bc+ac)^2\geq 3(a+b+c)\)
Do đó: \(P\geq ab+bc+ac+\frac{9}{2(a+b+c)}\geq 3\sqrt[3]{\frac{27}{8}}=\frac{9}{2}\)
Vậy \(P_{\min}=\frac{9}{2}\Leftrightarrow a=b=c=1\)
ap dung bdt cosi ta co : \(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge3\sqrt[3]{\dfrac{abc}{\left(abc\right)^2}}=3\) (1)
ta lai co \(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow\dfrac{9}{2\left(a+b+c\right)}=\dfrac{9\left(a+b+c\right)}{2\left(a+b+c\right)^2}\ge\dfrac{9.3}{2.3^2}=\dfrac{3}{2}\) (2)
tu (1) vs (2) \(\Rightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{9}{2\left(a+b+c\right)}\ge3+\dfrac{3}{2}=\dfrac{9}{2}\)
dau "=" xay ra khi \(a=b=c=1\)
xl ! may mk bi hu nen khong viet dau dc bn thong cam
\(\dfrac{a^2}{2b+c}+\dfrac{1}{9}.\left(2b+c\right)\ge2.\sqrt{\dfrac{a^2}{2b+c}.\dfrac{1}{9}\left(2b+c\right)}=\dfrac{2a}{3}\)
Tuong tu : \(\dfrac{b^2}{2c+a}+\dfrac{1}{9}\left(2c+a\right)\ge\dfrac{2b}{3}\)
\(\dfrac{c^2}{2a+b}+\dfrac{1}{9}\left(2a+b\right)\ge\dfrac{2c}{3}\)
=> P+\(\dfrac{2b+c+2c+a+2a+b}{9}\ge\dfrac{2}{3}\left(a+b+c\right)\)
=> P+\(\dfrac{3\left(a+b+c\right)}{9}\ge\dfrac{2}{3}\left(a+b+c\right)\)
=> P ≥ \(\dfrac{1}{3}\left(a+b+c\right)=\dfrac{2018}{3}\)
Ta có: \(\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{b}\)
\(\Rightarrow bc+ca=2ca\)
\(P=\dfrac{a+b}{2a-b}+\dfrac{c+b}{2c-b}=\dfrac{ac+bc}{2ca-bc}+\dfrac{ca+ab}{2ca-ab}\)
\(=\dfrac{ca+bc}{ab}+\dfrac{ca+ab}{bc}=\dfrac{c}{b}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{a}{c}=\dfrac{c+a}{b}+\dfrac{c}{a}+\dfrac{a}{c}\)
Ta có :
\(\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{4}{a+c}\left(\text{Svácxơ}\right)\)\(\Rightarrow c+a\ge2b\)
Áp dụng bđt cô si cho 2 số dương
\(\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{c}{a}.\dfrac{a}{c}}=2\)
\(\Rightarrow P\ge\dfrac{2b}{b}+2=4\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)