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4 tháng 8 2018

\(M=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{\left(1+1+1\right)^2}{a+b+c+3}\)

\(\ge\frac{3^2}{1+3}=\frac{9}{4}\)

=>MinM=9/4 khi a=b=c=1/3

4 tháng 8 2018

sai rồi

26 tháng 12 2016

Ta thấy:

\(\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)=\left(a+b+c\right)^2\le1\)

Áp dụng BĐT AM-GM ta có:

\(P\ge\left[\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)\right]\left(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\right)\)

\(\ge3\sqrt[3]{\left(a^2+2bc\right)\left(b^2+2ac\right)\left(c^2+2ab\right)}\cdot3\sqrt[3]{\frac{1}{a^2+2bc}\cdot\frac{1}{b^2+2ac}\cdot\frac{1}{c^2+2ab}}=9\)

Dấu "="xảy ra khi \(\left\{\begin{matrix}a+b+c=1\\a^2+2bc=b^2+2ac=c^2+2ab\end{matrix}\right.\)\(\Rightarrow a=b=c=\frac{1}{3}\)

Vậy \(Min_P=9\) khi \(a=b=c=\frac{1}{3}\)

NV
28 tháng 2 2020

\(A=\frac{1}{a+a+a+a+b+c}+\frac{1}{a+b+b+b+b+c}+\frac{1}{a+b+c+c+c+c}\)

\(\Rightarrow A\le\frac{1}{36}\left(\frac{4}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{4}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)

\(\Rightarrow A\le\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{6}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

29 tháng 5 2018

MÌnh nghĩ đề phải là tìm GTLN chứ

Ta có: \(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}=2\)

\(\Rightarrow\frac{1}{a+b+1}=\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\ge2\sqrt{\frac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}\)

Tương tự: \(\frac{1}{b+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(c+a\right)}{\left(a+b+1\right)\left(c+a+1\right)}}\)

                 \(\frac{1}{c+a+1}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)

Nhân lại ta có: \(\frac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\ge\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)

Dấu = khi a=b=c=1/4

21 tháng 2 2020

Áp dụng BĐT AM - GM ta có:

\(\frac{a+1}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}=a+1-\frac{b+ab}{2}\left(1\right)\)

Chứng minh tương tự ta có:

\(\frac{b+1}{c^2+1}\ge b+1-\frac{c+bc}{2}\left(2\right)\)

\(\frac{c+1}{a^2+1}\ge c+1-\frac{a+ca}{2}\left(3\right)\)

Từ: \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge\frac{a+b+c}{2}+3-\frac{ab+bc+ca}{2}\)

Lại có: \(a^2+b^2+c^2\ge ab+bc+ca\)

Hay: \(3\left(ab+bc+ac\right)\le\left(a+b+c\right)^2=3^2=9\)

Vì vậy: \(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge\frac{a+b+c}{2}+3-\frac{ab+bc+ca}{2}=\frac{3}{2}+3-\frac{9}{6}=3\)

\(\Rightarrow\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge3\)

\(\Rightarrow Min_P=3\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

21 tháng 2 2020

* Dũng kỹ thuật Cô-si ngược dấu

\(P=\left(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\right)+\left(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\)

+ \(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}=a-\frac{ab^2}{b^2+1}+b-\frac{bc^2}{c^2+1}+c-\frac{ca^2}{a^2+1}\)

\(\ge3-\left(\frac{ab^2}{2b}+\frac{bc^2}{2c}+\frac{ca^2}{2a}\right)=3-\frac{ab+bc+ca}{2}\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}\)

Dấu "=" \(\Leftrightarrow a=b=c=1\)

+ \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}=1-\frac{a^2}{a^2+1}+1-\frac{b^2}{b^2+1}+1-\frac{c^2}{c^2+1}\ge3-\left(\frac{a^2}{2a}+\frac{b^2}{2b}+\frac{c^2}{2c}\right)=3-\frac{a+b+c}{2}=\frac{3}{2}\)

Dấu "=" \(\Leftrightarrow a=b=c=1\)

Do đó: \(P\ge3\). Dấu "=" \(\Leftrightarrow a=b=c=1\)