Mai mk giup cho nhe

đơn giản : 

a + b + c = 1/a + 1/b + 1/c

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

Mà abc = 1 nên a + b + c = ab + bc + ac

cần chứng minh : ( a - 1 ) ( b - 1 ) ( c - 1 ) = 0

xét ( a - 1 )( b - 1 )( c - 1 ) = ( ab -a - b + 1 ) ( c - 1 )

= abc -ab - ac + a - bc + b + c - 1

= ( abc - 1 ) + ( a + b + c ) - ( ab + bc + ac )

= 0

\(x^5-x^4-x^3-x^2-x-2=0\)

\(\Leftrightarrow\left(x^5-2x^4\right)+\left(x^4-2x^3\right)+\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^4+x^3+x^2+x+1\right)=0\)

\(\Leftrightarrow x=2\)

Ta có:\(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\)

\(\Leftrightarrow2a^2+2b^2-\left(a+b\right)^2\ge0\)

\(\Leftrightarrow\) \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge1\)

\(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\)

Dấu "=" khi \(a=b=\frac{1}{2}\)

Áp dụng BĐT Bunhiacopxki ta có:

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

\(\Leftrightarrow2\left(a^2+b^2\right)\ge1\)

\(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow\frac{a}{1}=\frac{b}{1}\Leftrightarrow a=b=\frac{1}{2}\)

\(x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)=-2\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

Ta lại có: 

\(x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)^3=1\)

\(\Leftrightarrow x+y+z=1\)

Làm nốt

biến đổi tương đương đi, nhân tung ngoặc ra

\(\left(a+b\right)\left(a^4+b^4\right)\ge\left(a^2+b^2\right)\left(a^3+b^3\right)\)

\(\Leftrightarrow a^5+ab^4+a^4b+b^5\ge a^5+a^2b^3+a^3b^2+b^5\)

\(\Leftrightarrow ab^4+a^4b-a^2b^3-a^3b^2\ge0\)

\(\Leftrightarrow ab\left(a^3+b^3-ab^2-a^2b\right)\ge0\)

\(\Leftrightarrow a^3+b^3-ab^2-a^2b\ge0\)(Do ab > 0)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)Luôn đúng do a,b dương

Dấu "='' khi a = b