\(1+\dfrac{9}{3\left(ab+bc+ca\right)}\ge1+\dfrac{9}{\left(a+b+c\right)^2}\ge2\sqrt{\dfrac{9}{\left(a+b+c\right)^2}}=\dfrac{6}{a+b+c}\)

Lời giải:

Do \(3=ab+bc+ac\) nên ta có:

\(P=\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\)

\(=\frac{a^3}{b^2+ab+bc+ac}+\frac{b^3}{c^2+ab+bc+ac}+\frac{c^3}{a^2+ab+bc+ac}\)

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

Áp dụng BĐT AM-GM:

\(\frac{a^3}{(b+c)(b+a)}+\frac{b+c}{8}+\frac{b+a}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)

\(\frac{b^3}{(c+a)(c+b)}+\frac{c+a}{8}+\frac{c+b}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)

\(\frac{c^3}{(a+b)(a+c)}+\frac{a+b}{8}+\frac{a+c}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)

Cộng các BĐT trên vào và rút gọn:

\(\Rightarrow P+\frac{a+b+c}{2}\geq \frac{3}{4}(a+b+c)\)

\(\Rightarrow P\geq \frac{a+b+c}{4}(1)\)

Ta có một hệ quả quen thuộc của BĐT AM-GM đó là:

\((a+b+c)^2\geq 3(ab+bc+ac)\Leftrightarrow (a+b+c)^2\geq 9\)

\(\Rightarrow a+b+c\geq 3(2)\)

Từ \((1); (2)\Rightarrow P\geq \frac{3}{4}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

Lời giải:

\(\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c\)

\(\Leftrightarrow \frac{a^2+bc}{b+c}-c+\frac{b^2+ac}{a+c}-a+\frac{c^2+ab}{a+b}-b\geq 0\)

\(\Leftrightarrow \frac{a^2-c^2}{b+c}+\frac{b^2-a^2}{a+c}+\frac{c^2-b^2}{a+b}\geq 0\)

\(\Leftrightarrow a^2\left(\frac{1}{b+c}-\frac{1}{a+c}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\geq 0\)

\(\Leftrightarrow \frac{a^2(a-b)(a+b)+b^2(b-c)(b+c)+c^2(c-a)(c+a)}{(a+b)(b+c)(c+a)}\geq 0\)

\(\Leftrightarrow a^2(a^2-b^2)+b^2(b^2-c^2)+c^2(c^2-a^2)\geq 0\)

\(\Leftrightarrow a^4+b^4+c^4-(a^2b^2+b^2c^2+c^2a^2)\geq 0\)

\(\Leftrightarrow \frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{2}\geq 0\) (luôn đúng)

Do đó ta có đpcm

Dấu bằng xảy ra khi $a=b=c$

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

Áp dụng BĐT Cauchy-Schwarz:

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(b+c\right)\left(a+c\right)}}\le\dfrac{\dfrac{ab}{b+c}+\dfrac{ab}{a+c}}{2}\end{matrix}\right.\)

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

Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)

\(VT=\dfrac{a}{b\left(b^2+a\right)}+\dfrac{b}{c\left(c^2+b\right)}+\dfrac{c}{a\left(a^2+c\right)}\)

\(VT=\dfrac{a+b^2-b^2}{b\left(b^2+a\right)}+\dfrac{b+c^2-c^2}{c\left(c^2+b\right)}+\dfrac{c+a^2-a^2}{a\left(a^2+c\right)}\)

\(VT=\dfrac{1}{b}-\dfrac{b}{b^2+a}+\dfrac{1}{c}-\dfrac{c}{c^2+b}+\dfrac{1}{a}-\dfrac{a}{a^2+c}\)

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

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{b}{b^2+a}\le\dfrac{b}{2b\sqrt{a}}=\dfrac{1}{2\sqrt{a}}\)

Thiết lập tương tự và thu lại tao có

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\sqrt{\dfrac{1}{a}}\le\dfrac{\dfrac{1}{a}+1}{2}\)

Tương tự ta có

\(\sqrt{\dfrac{1}{b}}\le\dfrac{\dfrac{1}{b}+1}{2};\sqrt{\dfrac{1}{c}}\le\dfrac{\dfrac{1}{c}+1}{2}\)

Thu lại ta có

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3}{2}\right)\)

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)

\(\Rightarrow VT\ge\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\)

Áp dụng bất đẳng thức Cauchy dạng phân thức

\(\Rightarrow\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\ge\dfrac{3}{4}.\dfrac{9}{a+b+c}-\dfrac{3}{4}=\dfrac{3}{2}\)

\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b=c=1\)

Tách biểu thức như sau:

\(\left(\dfrac{a}{9}+\dfrac{b}{12}+\dfrac{c}{6}+\dfrac{8}{abc}\right)+\left(\dfrac{a}{18}+\dfrac{b}{24}+\dfrac{2}{ab}\right)+\left(\dfrac{b}{16}+\dfrac{c}{8}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{c}{6}+\dfrac{2}{ca}\right)+\left(\dfrac{13a}{18}+\dfrac{13b}{24}\right)+\left(\dfrac{13b}{48}+\dfrac{13c}{24}\right)\)