Bất đẳng thức cần chứng minh tương đương với:

\(\dfrac{a}{a+3\sqrt{bc}}+\dfrac{b}{b+3\sqrt{ca}}+\dfrac{c}{c+3\sqrt{ab}}\)

Ta áp dụng bất đẳng thức Cô si dạng \(2\sqrt{xy}\le x+y\) cho các căn thức ở mẫu, khi đó ta được:

\(\dfrac{a}{a+3\sqrt{bc}}+\dfrac{b}{b+3\sqrt{ca}}+\dfrac{c}{c+3\sqrt{ab}}\ge\) với biểu thức

\(\dfrac{2a}{2a+3b+3c}+\dfrac{2b}{3a+2b+3c}+\dfrac{2c}{3a+3b+2c}\)

Khi đó ta cần chứng minh: 

\(\dfrac{2a}{2a+3b+3c}+\dfrac{2b}{3a+2b+3c}+\dfrac{2c}{3a+3b+2c}\ge\dfrac{3}{4}\)

Đặt: \(\left\{{}\begin{matrix}x=2a+3b+3c\\y=3a+2b+3c\\z=3a+3b+2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a=\dfrac{1}{4}\left(3y+3z-5x\right)\\2b=\dfrac{1}{4}\left(3z+3x-5y\right)\\2c=\dfrac{1}{4}\left(3x+3y-5z\right)\end{matrix}\right.\)

Khi đó đẳng thức trên được viết lại thành:

\(\dfrac{3y+3z-5x}{4x}+\dfrac{3z+3x-5y}{4y}+\dfrac{3x+3y-5z}{4z}\ge\dfrac{3}{4}\)

Hay: \(3\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{z}{x}\right)-15\ge3\)

Bất đẳng thức cuối cùng luôn đúng theo bất đẳng thức Cô si.

Vậy bất đẳng thức được chứng minh. Đẳng thức xảy ra khi và chỉ khi \(a=b=c\)

Đặt \(x=\sqrt{a};y=\sqrt{b};z=\sqrt{c}\)

Khi đó bđt đã tro chở thành:

\(\dfrac{yz}{x^2+3yz}+\dfrac{zx}{y^2+3zx}+\dfrac{xy}{z^2+3xy}\le\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{yz}{x^2+3yz}+\dfrac{1}{3}-\dfrac{zx}{y^2+3zx}+\dfrac{1}{3}-\dfrac{xy}{z^2+3xy}\ge1-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{x^2}{x^2+3yz}+\dfrac{y^2}{y^2+3zx}+\dfrac{z^2}{z^2+3xy}\ge\dfrac{3}{4}\) (đpcm)

Áp dụng AM-GM:

\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)

Tương tự rồi cộng lại:

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

\(=\frac{3}{2}\)

Đẳng thức xảy ra tại a=b=c=¹/₃

Do \(a+b+c=1\)  nên :

\(VT=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\frac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\frac{ca}{b\left(a+b+c\right)+ac}}\)

\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)

Áp dụng BĐT AM - GM :
\(\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)

\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right)\)

\(\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)

Cộng theo vế :
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\left(đpcm\right)\)

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

Chúc bạn học tốt !!!

\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)

Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)

\(\frac{\sqrt{ab}}{c+2\sqrt{ab}}=\frac{1}{2}\left(\frac{x+2\sqrt{xy}-z}{z+2\sqrt{xy}}\right)=\frac{1}{2}\left(1-\frac{z}{z+2\sqrt{xy}}\right)\le\frac{1}{2}\left(1-\frac{z}{x+y+z}\right)\)

Tương tự \(\frac{\sqrt{yz}}{x+2\sqrt{yz}}\le\frac{1}{2}\left(1-\frac{x}{x+y+z}\right)\);\(\frac{\sqrt{xz}}{y+2\sqrt{xz}}\le\frac{1}{2}\left(1-\frac{y}{x+y+z}\right)\)

Cộng vế theo vế ta được \(\frac{\sqrt{xy}}{z+2\sqrt{xy}}+\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}\le\frac{1}{2}\left(3-1\right)=1\)

bạn cho mình hỏi x,y,z là j vậy bạn

ok, mik hiểu r, cảm ơn bạn nhiều

\(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+a^2\ge2ca.\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Leftrightarrow ab+bc+ca\le\frac{3^2}{3}=3\)

Khi đó \(c^2+3\ge c^2+ab+bc+ca=\left(b+c\right)\left(a+c\right)\Leftrightarrow\sqrt{c^2+3}\ge\sqrt{b+c}\sqrt{a+c}\)

     \(a^2+3\ge a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\Leftrightarrow\sqrt{a^2+c}\ge\sqrt{\left(a+b\right)}\sqrt{a+c}\)

\(b^2+3\ge b^2+ab+bc+ca=\left(a+b\right)\left(b+c\right)\Leftrightarrow\sqrt{b^2+3}\ge\sqrt{a+b}\sqrt{b+c}\)

\(\Rightarrow\frac{ab}{\sqrt{c^2+3}}+\frac{bc}{\sqrt{a^2+3}}+\frac{ca}{\sqrt{b^2+3}}\le\frac{ab}{\sqrt{b+c}\sqrt{a+c}}+\frac{bc}{\sqrt{a+b}\sqrt{a+c}}+\frac{ca}{\sqrt{a+b}\sqrt{b+c}}\)*

áp dụng bđt Cauchy ngược dấu 

\(\sqrt{\frac{1}{a+b}}.\sqrt{\frac{1}{a+c}}\le\frac{\frac{1}{a+b}+\frac{1}{a+c}}{2}\Leftrightarrow\frac{2}{\sqrt{a+b}\sqrt{a+c}}\le\frac{1}{a+b}+\frac{1}{a+c}\)

\(\Leftrightarrow\frac{2bc}{\sqrt{a+b}\sqrt{a+c}}\le\frac{bc}{a+b}+\frac{bc}{a+c}\)

Chứng minh tương tự \(\frac{2ab}{\sqrt{a+c}\sqrt{b+c}}\le\frac{ab}{a+c}+\frac{ab}{b+c}\)

                                   \(\frac{2ca}{\sqrt{b+c}\sqrt{a+b}}\le\frac{ca}{b+c}+\frac{ca}{a+b}\)

Kết hợp với * ta có 

\(\frac{2ab}{\sqrt{c^2+3}}+\frac{2bc}{\sqrt{a^2+3}}+\frac{2ca}{\sqrt{b^2+3}}\le\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+c}+\frac{bc}{a+b}+\frac{ca}{a+b}+\frac{ca}{b+c}\)

\(\Leftrightarrow2\left(\frac{ab}{\sqrt{c^2+3}}+\frac{bc}{\sqrt{a^2+3}}+\frac{ca}{\sqrt{b^2+3}}\right)=\frac{bc+ca}{a+b}+\frac{ab+bc}{a+c}+\frac{ab+ca}{b+c}=a+b+c\)

\(\Leftrightarrow\frac{ab}{\sqrt{c^2+3}}+\frac{bc}{\sqrt{a^2+3}}+\frac{ca}{\sqrt{b^2+3}}\le\frac{a+b+c}{2}=\frac{3}{2}.\)

nhầm xíu dòng thứ 2 từ dưới lên 

\(2\left(...\right)\ge\frac{ab}{..}...\)=...

mk lại nhầm nữa đổi từ dấu \(=thành\le\)

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

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

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

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