Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)

\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)

\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)

\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)

Xảy ra khi \(a=b=c\)

\(BĐT\Leftrightarrow\sum\left(\dfrac{1}{a}-\dfrac{b+c}{a^2+bc}\right)\ge0\)

\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\)

Giả sử \(a\ge b\ge c\)thì

\(\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\).vậy nên chỉ cần chứng minh

\(\dfrac{\left(b-c\right)\left(b-a\right)}{b\left(b^2+ac\right)}+\dfrac{\left(c-a\right)\left(c-b\right)}{c\left(c^2+ab\right)}\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\dfrac{b-a}{b\left(b^2+ac\right)}+\dfrac{a-c}{c\left(c^2+ab\right)}\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\left(b-a\right)\left(c^3+abc\right)+\left(a-c\right)\left(b^3+abc\right)\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)^2\left(b+c\right)\left(ab+ac-bc\right)\ge0\)( đúng vì \(a\ge b\ge c\))

Vậy BĐT được chứng minh.

Dấu = xảy ra khi a=b=c

\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:

\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+a}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\)

\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)

\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)

Cộng vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

nhầm mọi người ơi chứng minh cho mình <=\(\dfrac{3}{\sqrt{2}}\)

\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)

\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)

\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)

Áp dụng BĐT quen thuộc:

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{8}{9}\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\)

Ta chỉ cần chứng minh:

\(\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\le\dfrac{ab+bc+ca}{abc}\)

\(\Leftrightarrow4\left(ab+bc+ca\right)^2\ge9abc+9abc\left(ab+bc+ca\right)\)

Do \(3\left(ab+bc+ca\right)^2\ge9abc\left(a+b+c\right)=9abc\)

Nên ta chỉ cần chứng minh:

\(\left(ab+bc+ca\right)^2\ge9abc\left(ab+bc+ca\right)\)

\(\Leftrightarrow ab+bc+ca\ge9abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)

Hiển nhiên đúng do \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)

Ta có: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\end{matrix}\right.\Leftrightarrow\left(1-a\right)\left(1-b\right)\ge0\)

\(\Rightarrow1-b-a+ab\ge0\Leftrightarrow1+ab\ge a+b\)(1)

Tiếp tục chứng minh ta được: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1\ge c\\ab\ge0\end{matrix}\right.\)(2)

Cộng theo vế pt(1) với pt(2) ta được:

\(1+ab+1+ab\ge a+b+c+0\)

\(\Rightarrow2\left(ab+1\right)\ge a+b+c\)

Nên: \(\dfrac{c}{ab+1}=\dfrac{2c}{2\left(ab+1\right)}\le\dfrac{2c}{a+b+c}\)

Chứng minh tương tự suy ra đpcm

Câu hỏi của Phạm Quốc Anh - Toán lớp 7 - Học toán với OnlineMath