Với a; b dương chứ nhỉ, nằm dưới mẫu thêm điều kiện khác 0, mà không âm + khác 0 thì nó là dương còn gì?

\(\Leftrightarrow\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\le\sqrt[3]{2\left(\frac{a}{b}+\frac{b}{a}+2\right)}\)

\(\Leftrightarrow\left(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\right)^3\le2\left(\frac{a}{b}+\frac{b}{a}+2\right)\)

Đặt \(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}=x\ge2\) BĐT tương đương:

\(x^3\le2\left(x^3-3x+2\right)\)

\(\Leftrightarrow x^3-6x+4\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-2\right)\ge0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+x+x-2\right]\ge0\) (luôn đúng)

Vậy BĐT được chứng minh, dấu "=" xảy ra khi \(x=2\Leftrightarrow a=b\)

Áp dụng BĐT Cauchy-Schwarz ta có:

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

\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\). Thiếp lập 2 BĐT còn lại:

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

Cộng theo vế 3 BĐT trên ta có:

\(A\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)

Xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Ta có \(a+b+c\le\sqrt{3}\)

\(\Rightarrow\left(a+b+c\right)^2\le3\)

\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\le1\)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ac\)

\(\Rightarrow1\ge ab+bc+ac\)

\(\Rightarrow\left\{\begin{matrix}1+a^2\ge a^2+ab+bc+ac\\1+b^2\ge b^2+ab+bc+ac\\1+c^2\ge c^2+ab+bc+ac\end{matrix}\right.\)

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

\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{1+a^2}}\le\frac{a}{\sqrt{a^2+ab+bc+ac}}\\\frac{b}{\sqrt{1+b^2}}\le\frac{b}{\sqrt{b^2+ab+bc+ac}}\\\frac{c}{\sqrt{1+c^2}}\le\frac{c}{\sqrt{c^2+ab+bc+ac}}\end{matrix}\right.\)

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

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

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

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

Áp dụng bất đẳng thức Cauchy ngược dấu cho 2 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}\sqrt{\left(a+b\right)\left(a+c\right)}\ge\frac{2a+b+c}{2}\\\sqrt{\left(a+b\right)\left(b+c\right)}\ge\frac{a+2b+c}{2}\\\sqrt{\left(c+a\right)\left(c+b\right)}\ge\frac{a+b+2c}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{2a}{2b+b+c}\\\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{2b}{a+2b+c}\\\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{2c}{a+b+2c}\end{matrix}\right.\)

\(\Rightarrow\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\)

Chứng minh rằng: \(2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\le\frac{3}{2}\)

\(\Leftrightarrow\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\le\frac{3}{4}\)

Áp dụng bất đẳng thức \(\frac{1}{a+b}\ge\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\frac{a}{2a+b+c}=\frac{a}{a+c+a+b}\le\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)

\(\Rightarrow\frac{b}{a+2b+c}=\frac{b}{a+b+b+c}\le\frac{b}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)

\(\Rightarrow\frac{c}{a+b+2c}=\frac{c}{a+c+b+c}\le\frac{c}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)

\(\Rightarrow VT\le\frac{a}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}+\frac{b}{4\left(a+b\right)}+\frac{b}{4\left(b+c\right)}+\frac{c}{4\left(a+c\right)}+\frac{c}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\frac{a}{4\left(a+b\right)}+\frac{b}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}+\frac{c}{4\left(a+c\right)}+\frac{b}{4\left(b+c\right)}+\frac{c}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\left(đpcm\right)\)

\(\Rightarrow2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\le\frac{3}{2}\)

\(\Rightarrow\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{3}{2}\)

Vậy \(\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{3}{2}\left(đpcm\right)\)

Lời giải khác:

Áp dụng bđt Cauchy-Schwarz:

\((a^2+1)(1+3)\geq (a+\sqrt{3})^2\)\(\Rightarrow \frac{a}{\sqrt{a^2+1}}\leq \frac{2a}{a+\sqrt{3}}\)

Thực hiện tương tự với các phân thức còn lại:

\(\Rightarrow \frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\leq 2\left ( \frac{a}{a+\sqrt{3}}+\frac{b}{b+\sqrt{3}}+\frac{c}{c+\sqrt{3}} \right )=2A\) $(1)$

Lại có:

\(\)\(A=\left ( 1-\frac{\sqrt{3}}{a+\sqrt{3}} \right )+\left ( 1-\frac{\sqrt{3}}{b+\sqrt{3}} \right )+\left ( 1-\frac{\sqrt{3}}{c+\sqrt{3}} \right )=3-\sqrt{3}\left ( \frac{1}{a+\sqrt{3}}+\frac{1}{b+\sqrt{3}}+\frac{1}{c+\sqrt{3}} \right )\)

Cauchy-Schwarz kết hợp với \(a+b+c\leq \sqrt{3}\):

\(A\leq 3-\frac{9\sqrt{3}}{a+b+c+3\sqrt{3}}\leq 3-\frac{9\sqrt{3}}{4\sqrt{3}}=\frac{3}{4}\) $(2)$

Từ \((1),(2)\Rightarrow \text{VT}\leq 2A\leq \frac{3}{2}\) (đpcm)

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

Cách khác nữa:

Nhớ là \(f\left(x\right)=\frac{x}{\sqrt{x^2+1}}\) là 1 hàm lõm khi x>0, điều này xảy ra khi

\(f''(x)=-\dfrac{3x}{(x^2+1)^{\frac{5}{2}}}<0\). giờ thì sử dụng BĐT jensen

\(f\left(a\right)+f\left(b\right)+f\left(c\right)\le3f\left(\frac{a+b+c}{3}\right)=3f\left(\frac{\sqrt{3}}{3}\right)=\frac{3}{2}\left(a+b+c=\sqrt{3}\right)\)

Đạt dc GTLN khi \(a=b=c\).

Vì a,b,c là số thực dương nên \(\sqrt{a^2}=a;\sqrt{b^2}=b;\sqrt{c^2}\)=c. Vậy ta có

\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\)=\(\frac{a}{a+1}-1+\frac{b}{b+1}-1\)+\(\frac{c}{c+1}-1+3\) 

=3-(  \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\)) =A

ta có bdt  \(9\le\left(a+1+b+1+c+1\right)\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)(dễ dàng chứng mình bằng bdt cosi).

=>\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\)\(\frac{9}{3+\sqrt{3}}\)=> A\(\le3-\frac{9}{3+\sqrt{3}}=\frac{3\sqrt{3}}{3+\sqrt{3}}=\frac{3}{\sqrt{3}+1}\)

dấu = khi a=b=c=\(\frac{\sqrt{3}}{3}\)