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Chính bài của em:
Cho \(a,b,c\ge1\). CMR: \(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)+2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}... - Hoc24
do \(a,b,c\ge1\)\(=>\left\{{}\begin{matrix}b+c\ge2\\c+a\ge2\\a+b\ge2\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}a\left(b+c\right)\ge2a\\b\left(c+a\right)\ge2b\\c\left(a+b\right)\ge2c\end{matrix}\right.\)
\(=>\) biểu thức đề bài cho\(\ge2\left(a+b+c+\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)
\(2\left(1+1+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\right)=9\)
dấu= xảy ra<=>a=b=c=1
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)
\(=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+3\)
Áp dụng BĐT Cô - si cho 2 số không âm:
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)
Suy ra:
\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+3\ge2+2+2+3=9\)
Dấu "=" xảy ra khi: a = b = c
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(\dfrac{9}{a+b+c}\)
⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}\left(a+b+c\right).\dfrac{9}{a+b+c}\)
⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}9\)
\("="\text{⇔}a=b=c\)
\(ab+1\le b\Rightarrow a+\dfrac{1}{b}\le1\)
Đặt \(\left(a;\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow x+y\le1\)
\(P=x+\dfrac{1}{x^2}+y+\dfrac{1}{y^2}=\left(\dfrac{x}{2}+\dfrac{x}{2}+\dfrac{1}{16x^2}\right)+\left(\dfrac{y}{2}+\dfrac{y}{2}+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(P\ge3\sqrt[3]{\dfrac{x^2}{64x^2}}+3\sqrt[3]{\dfrac{y^2}{64y^2}}+\dfrac{15}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(P\ge\dfrac{3}{2}+\dfrac{15}{32}\left(\dfrac{4}{x+y}\right)^2\ge\dfrac{3}{2}+\dfrac{15}{32}.\left(\dfrac{4}{1}\right)^2=9\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\) hay \(\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)
Viết gọn lại, ta cần chứng minh:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\right)\)
\(\Leftrightarrow\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{a+b}{ab}}\right)=\sum\dfrac{4ab}{a+b}\)
Thật vậy, ta có:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum\left(2\sqrt{\left(a+b\right).\dfrac{1}{4}}\right)^2=\sum a+b\)
Vậy ta cần chứng minh:
\(\sum a+b\ge\sum\dfrac{4ab}{a+b}\Leftrightarrow\sum\left(a+b\right)^2\ge\sum4ab\Leftrightarrow\sum\left(a-b\right)^2\ge0\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c
Áp dụng BĐT cauchy ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\dfrac{1}{abc}}=9\sqrt[3]{abc\cdot\dfrac{1}{abc}}=9\)
Dấu \("="\Leftrightarrow a=b=c\)