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?

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

\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2a}{2b+a+b}+\dfrac{2b}{2c+b+c}+\dfrac{2c}{2a+a+c}\)

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

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

\(\Rightarrow P\ge\dfrac{3\sqrt{2}}{2}\) (đpcm)

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

làm tương tự với \(\dfrac{b}{\sqrt{bc+c^2}};\dfrac{c}{\sqrt{ca+a^2}}\)

\(=>P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\)

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

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

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

dấu"=" xảy ra<=>a=b=c

a)Svac-so:

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

b)\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)

\(\Leftrightarrow\dfrac{1}{a^2+1}-\dfrac{1}{ab+1}+\dfrac{1}{b^2+1}-\dfrac{1}{ab+1}\ge0\)

\(\Leftrightarrow\dfrac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{b\left(a-b\right)}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b}{\left(b^2+1\right)\left(ab+1\right)}-\dfrac{a}{\left(a^2+1\right)\left(ab+1\right)}\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b\left(a^2+1\right)-a\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(\dfrac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(\dfrac{ab\left(a-b\right)-\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\cdot\dfrac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)(luôn đúng)

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)

Ta có:

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

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

Dấu = xảy ra khi \(a=b\)

Bài 1:

\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0

Áp dụng BĐT Chauchy cho 2 số không âm, ta có:

\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge c\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2c\)

\(\dfrac{ac}{b}+\dfrac{ab}{c}=a\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge a\sqrt{\dfrac{c}{b}.\dfrac{b}{c}}=2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}=b\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge b\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2b\)

Cộng vế theo vế ta được:

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

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

\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)

Dấu "=" xảy ra khi \(a=b=c\)

Ta chứng minh bđt phụ \(x^2+y^2+z^2\ge xy+yz+zx\forall x,y,z>0\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\Leftrightarrow x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\ge0\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)\(\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\left(1\right)\)

Áp dụng bđt Cô-si vào các số a,b,c dương :

\(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}\cdot ab}=2\sqrt{a^4}=2a^2\)

Chứng minh tương tự ta được:

\(\dfrac{b^3}{c}+bc\ge2b^2;\dfrac{c^3}{a}+ca\ge2c^2\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2a^2+2b^2+2c^2\ge2ab+2bc+2ca\) (do áp dụng (1)) \(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(ab+bc+ca\right)-\left(ab+bc+ca\right)=ab+bc+ca\)

Dấu = xảy ra \(\Leftrightarrow a=b=c\)