AM-GM ngược dấu như sau:

\(\dfrac{a^3}{a^2+ab+b^2}=a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\dfrac{ab\left(a+b\right)}{3ab}=\dfrac{2a-b}{3}\)

Tương tự ta cho 2 BĐT còn lại ta cũng có:

\(\dfrac{b^3}{b^2+bc+c^2}\ge\dfrac{2b-c}{3};\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{2c-a}{3}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\dfrac{2a-b}{3}+\dfrac{2b-c}{3}+\dfrac{2c-a}{3}=\dfrac{a+b+c}{3}=VP\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)

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

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

Dễ thấy :

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

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

Vậy cần chứng minh

\(\dfrac{a^2+b^2+c^2}{a+b+c}\ge\dfrac{a+b+c}{3}\Leftrightarrow\left(a+b+c\right)^2\ge3\left(a^2+b^2+c^2\right)\) (luôn đúng)

Chứng minh : \(\left(x^2+y^2+z^2\right)^2\ge3\left(x^3y+y^3z+z^3x\right)\)

\(\Leftrightarrow\dfrac{1}{2}\left(\left(x^2-y^2-xy-xz+2yz\right)^2+\left(y^2-z^2-yz-xy+2xz\right)^2+\left(z^2-x^2-xz-yz+2xy\right)^2\right)\ge0\)

Áp dụng BĐT AM-GM ta có:

\(\dfrac{a}{ab+1}=a-\dfrac{a^2b}{ab+1}\ge a-\dfrac{a^2b}{2\sqrt{ab}}=a-\dfrac{\sqrt{a^3b}}{2}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\dfrac{b}{bc+1}\ge b-\dfrac{\sqrt{b^3c}}{2};\dfrac{c}{ca+1}\ge c-\dfrac{\sqrt{c^3a}}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge3-\dfrac{1}{2}\left(\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)

Xảy ra khi \(a=b=c=1\)

Ta có BĐT \(abc\le\left(\dfrac{a+b+c}{3}\right)^3=1\)

Và \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=3\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a}{ab+1}+\dfrac{b}{bc+1}+\dfrac{c}{ca+1}\)

\(\ge\dfrac{1}{b+bc}+\dfrac{1}{b+ac}+\dfrac{1}{c+ab}\)

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

Xảy ra khi \(a=b=c=1\)

áp dụng BĐT cô si cho 2 số ta có

\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{bc}{a}.\dfrac{ac}{b}}\)

⇔\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{c^2}=2c\)

TT ta có \(\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2b\)

cộng từng vế 3 BĐT trên

\(2\left(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\right)\ge2\left(a+b+c\right)\)

⇔ \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\) (đpcm)

e)

\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

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

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

=> ĐPCM

BPT?

a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)

\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)

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

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

b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)

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

\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)

b)

Áp dụng BĐT Cauchy Shwarz, ta có:

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2\)

Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:

\(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\dfrac{\left(a+b+c\right)^3}{9}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\dfrac{\left(a+b+c\right)}{3}\times\left(a^2+b^2+c^2\right)}\)

\(=\dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}\) (đpcm)

Dấu "=" xảy ra khi a = b = c.

Ta có: \(a^2+b^2+c^2\ge ab+bc+ca\ge\sqrt[]{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

Do đó:

\(VT\le\dfrac{2a^3}{2\sqrt{a^6bc}}+\dfrac{2b^3}{2\sqrt{b^6ac}}+\dfrac{2c^3}{2\sqrt{c^3ab}}=\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}=\dfrac{\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}{abc}\)

\(\le\dfrac{a^2+b^2+c^2}{abc}=\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\)

Dấu "=" xảy ra khi \(a=b=c=1\)