c=c.1 thay 1 bằng a+b+c xong cô si

\(3=a+b+c\ge3\sqrt[3]{abc}\Rightarrow abc\le1\)

BĐT tương đương:

\(3\left(ab+bc+ca\right)\ge abc\left[\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+6\right]\)

\(\Leftrightarrow3\left(ab+bc+ca\right)\ge abc\left[15-2\left(ab+bc+ca\right)\right]\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(2abc+3\right)\ge15abc\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\left(2abc+3\right)^2\ge225\left(abc\right)^2\)

Do \(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)=9abc\)

Nên ta chỉ cần chứng minh:

\(\left(2abc+3\right)^2\ge25abc\)

\(\Leftrightarrow\left(1-abc\right)\left(9-4abc\right)\ge0\) (luôn đúng với \(0< abc\le1\))

Dấu "=" xảy ra khi \(a=b=c=1\)

Đặt \(\left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)=\left(x,y,z\right)\) với x, y, z > 0 thì ta có \(x+y+z=1\).

Đặt biểu thức ở VT là A. Ta có: 

\(A=\sqrt{\dfrac{b^2+2a^2}{a^2b^2}}+\sqrt{\dfrac{c^2+2b^2}{b^2c^2}}+\sqrt{\dfrac{a^2+2c^2}{c^2a^2}}=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\).

Ta có bất đẳng thức \(\sqrt{a_1^2+a_2^2}+\sqrt{a_3^2+a_4^2}\ge\sqrt{\left(a_1+a_3\right)^2+\left(a_2+a_4\right)^2}\).

Đây là bđt Mincopxki cho hai bộ số thực và dễ dàng cm bằng biến đổi tương đương.

Do đó \(A\ge\sqrt{\left(x+y\right)^2+\left(\sqrt{2}y+\sqrt{2}z\right)^2}+\sqrt{z^2+2x^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\sqrt{2}y+\sqrt{2}z+\sqrt{2}x\right)^2}=\sqrt{1+2}=\sqrt{3}=VP\).

Đẳng thức xảy ra khi a = b = c = 3.

Vậy...

Tương tự: \(GT\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

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

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

\(VT\ge\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)

\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)

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

Tính M = ab + bc + ca/ a² + b² + c²

^{\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)}

^{\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)}

^{\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)}

^{\(\Rightarrow\hept{\begin{cases}\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}=\frac{1}{c}\\\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\Rightarrow\frac{1}{b}=\frac{1}{a}\\\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{a}=\frac{1}{c}=\frac{1}{a}\end{cases}}\)}

^{\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)}

\(\Rightarrow M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{1.1+1.1+1.1}{1^2+1^2+1^2}=\frac{3}{3}=1\)

Ta có \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

Mà \(a,b,c \ne0\) => \(ab,bc,ca \ne0\)

=> \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

=> \(\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)

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

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

=> \(a=b=c\)

Thay vào M ta có : \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a.a+a.a+a.a}{a^2+a^2+a^2}=\frac{3a^2}{3a^2}=1\)

 Vậy \(M=1\)

Ta có \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

Mà \(a,b,c \ne0\) => \(ab,bc,ca \ne0\)

=> \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

=> \(\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)

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

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

=> \(a=b=c\)

Thay vào M ta có : \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a.a+a.a+a.a}{a^2+a^2+a^2}=\frac{3a^2}{3a^2}=1\)

 Vậy \(M=1\)

Ta có : \(ab+bc+ca=0\)

<=> \(abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\left(\text{vì }a;b;c\ne0\right)\)

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

<=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}.\left(-\frac{1}{c}\right)\left(\text{vì }\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\right)\)

<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Khi đó \(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)

\(a^4+b^4+b^4+b^4\ge4\sqrt[4]{a^4b^{12}}=4ab^3\)

Tương tự:

\(b^4+3c^4\ge4bc^3\) ; \(c^4+3a^4\ge4ca^3\)

Cộng vế:

\(M\le a^4+b^4+c^4=1\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt[4]{3}}\)