\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(a-1\right)\left(bc-b-c+1\right)\)

\(=abc-\left(ab+bc+ca\right)+a+b+c-1\)

\(=abc-abc+1-1=0\) (đpcm)

Chắc là a;b;c hết chứ?

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

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

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

cho x,y,z>0 ,x+y+z=1 chu nhi?

\(\Rightarrow\dfrac{x}{x+y+z+y-x}=\dfrac{x}{2y+z}\)

\(\Rightarrow\dfrac{y}{1+z-y}=\dfrac{y}{x+y+z+z-y}=\dfrac{y}{2z+x}\)

\(\Rightarrow\dfrac{z}{1+x-z}=\dfrac{z}{x+y+z+x-z}=\dfrac{z}{2x+y}\)

\(\Rightarrow A=\dfrac{x}{2y+z}+\dfrac{y}{2z+x}+\dfrac{z}{2x+y}=\dfrac{x^2}{2xy+xz}+\dfrac{y^2}{2zy+xy}+\dfrac{z^2}{2xz+xz}\ge\dfrac{\left(x+y+z\right)^2}{3\left(xy+yz+xz\right)}=1\)

dau"=" xay ra<=>x=y=z=1/3

Sửa đề: 1+a^2;1+b^2;1+c^2

\(\dfrac{a}{\sqrt{1+a^2}}=\dfrac{a}{\sqrt{a^2+ab+c+ac}}=\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}< =\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

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

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

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

\(abc=1\) nên tồn tại các số dương x;y;z sao cho \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

BĐT cần chứng minh tương đương:

\(\dfrac{y}{x+2y}+\dfrac{z}{y+2z}+\dfrac{x}{z+2x}\le1\)

\(\Leftrightarrow\dfrac{2y}{x+2y}-1+\dfrac{2z}{y+2z}-1+\dfrac{2x}{z+2x}-1\le2-3\)

\(\Leftrightarrow\dfrac{x}{x+2y}+\dfrac{y}{y+2z}+\dfrac{z}{z+2x}\ge1\)

Điều này đúng do:

\(VT=\dfrac{x^2}{x^2+2xy}+\dfrac{y^2}{y^2+2yz}+\dfrac{z^2}{z^2+2xz}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=1\)

Áp dụng bất đẳng thức Cauchy:

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

CMTT: \(b\sqrt{a-1}\le\dfrac{ab}{2}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le ab\left(đpcm\right)\)

\(ĐTXR\Leftrightarrow a=b=1\)

\(Áp\ dụng\ BĐT\ AM - GM,\ ta\ có: \\\sum\dfrac{1}{a^2+2b^2+3}=\sum\dfrac{1}{(a^2+b^2)+(b^2+1)+2}\le\sum\dfrac{1}{2ab+2b+2} \\=\dfrac{1}{2}\sum\dfrac{1}{ab+b+1}=\dfrac{1}{2}.1=\dfrac{1}{2} \\Đẳng\ thức\ xảy\ ra\ khi\ a=b=c=1.\)

Giải giúp tui trong 5 phút ik

-Đề sai.

Nếu b=0 thì chắc A sẻ =1

Theo đó =>a²+b²=1²+0²⁼1+0=1

1. Phải là \((a+b+c)^{\color{red}{2}}=3(ab+bc+ac)\) chứ nhỉ?
VD: Với \(a=b=c=1\) thì \((a+b+c)^3=27\ne 3(ab+bc+ac)=9\) !!!