\(a+b+c=1=>\left\{{}\begin{matrix}1-a=b+c\\1-b=a+c\\1-c=a+b\\\end{matrix}\right.\)

\(=>A=\left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right)=\left(\dfrac{1-a}{a}\right)\left(\dfrac{1-b}{b}\right)\left(\dfrac{1-c}{c}\right)\)

\(=\left(\dfrac{b+c}{a}\right)\left(\dfrac{a+c}{b}\right)\left(\dfrac{a+b}{c}\right)\)

bbđt AM-GM

\(=>A\ge\dfrac{2\sqrt{bc}.2\sqrt{ac}.2\sqrt{ab}}{abc}=\dfrac{8abc}{abc}=8\left(đpcm\right)\)

dấu"=" xảy ra<=>\(a=b=c=\dfrac{1}{3}\)

Đặt vế trái BĐT cần chứng minh là P

Ta có:

\(P=\left(\dfrac{a+b+c}{a}-1\right)\left(\dfrac{a+b+c}{b}-1\right)\left(\dfrac{a+b+c}{c}-1\right)\)

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\ge\dfrac{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{abc}=8\) (đpcm)

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

lỗi rồi bạn nhé

\(ab+bc+ca\le1\)

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

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

\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}+\dfrac{\dfrac{b}{a+b}+\dfrac{b}{b+c}}{2}+\dfrac{\dfrac{c}{b+c}+\dfrac{c}{a+c}}{2}=\dfrac{3}{2}\left(đpcm\right)\)

\(dấu"="\Leftrightarrow a=b=c=\sqrt{\dfrac{1}{3}}\)

Có \(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\ge9\)

\(\Leftrightarrow\dfrac{a+1}{a}.\dfrac{b+1}{b}\ge9\)

\(\Leftrightarrow ab+a+b+1\ge9ab\) ( vì ab >0)

\(\Leftrightarrow a+b+1\ge8ab\)

\(\Leftrightarrow2\ge8ab\) \(\left(a+b=1\right)\)

\(\Leftrightarrow1\ge4ab\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\) \(\left(a+b=1\right)\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng)

\(\Leftrightarrowđpcm\)

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

\(A=\frac{c}{abc+ac+c}+\frac{ac}{abc\cdot c+abc+ac}+\frac{1}{ac+c+1}\)

\(A=\frac{c}{ac+c+1}+\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}\)

\(A=\frac{ac+c+1}{ac+c+1}\)

\(A=1\)

Với a;b;c dương:

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)-\sqrt[3]{abc}.\sqrt[3]{ab.bc.ca}\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\dfrac{1}{3}\left(a+b+c\right).\dfrac{1}{3}\left(ab+bc+ca\right)\)

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

Đặt vế trái BĐT là P, ta có:

\(\dfrac{ab}{1-c^2}=\dfrac{ab}{\left(1-c\right)\left(1+c\right)}=\dfrac{ab}{\left(a+b\right)\left(a+c+b+c\right)}=\dfrac{ab}{\sqrt{a+b}.\sqrt{a+b}\left(a+c+b+c\right)}\)

\(\le\dfrac{ab}{\sqrt[]{2\sqrt[]{ab}}.\sqrt[]{a+b}.2\sqrt[]{\left(a+c\right)\left(b+c\right)}}=\dfrac{\sqrt[4]{\left(ab\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Tương tự:

\(\dfrac{bc}{1-a^2}\le\dfrac{\sqrt[4]{\left(bc\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\dfrac{ca}{1-b^2}\le\dfrac{\sqrt[4]{\left(ca\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Cộng vế:

\(P\le\dfrac{\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}}{2\sqrt[]{2}.\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Nên ta chỉ cần chứng minh:

\(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\le\dfrac{3}{2\sqrt[]{2}}\sqrt[]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(\Leftrightarrow\left(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\right)^2\le\dfrac{9}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

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

Nên ta chỉ cần chứng minh:

\(\left(\sqrt[4]{\left(ab\right)^3}+\sqrt[4]{\left(bc\right)^3}+\sqrt[4]{\left(ca\right)^3}\right)^2\le\left(a+b+c\right)\left(ab+bc+ca\right)\)

Thật vậy:

\(\left(\sqrt[4]{ab}.\sqrt[]{ab}+\sqrt[4]{bc}.\sqrt[]{bc}+\sqrt[4]{ca}.\sqrt[]{ca}\right)^2\le\left(\sqrt[]{ab}+\sqrt[]{bc}+\sqrt[]{ca}\right)\left(ab+bc+ca\right)\)

\(\le\left(a+b+c\right)\left(ab+bc+ca\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

BĐT tương đương

\(\dfrac{a+c}{b+c}-\dfrac{a}{b}>0\Leftrightarrow\dfrac{ab+bc-ab-ac}{b\left(b+c\right)}>0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)}{b\left(b+c\right)}>0\)\(\Leftrightarrow b-a>0\Leftrightarrow b>a\Leftrightarrow\dfrac{a}{b}< 1\)(đúng vì GT)