TỰ NGHĨ NHAK

`Answer:`

a. Ta có: \(a^2+b^2-2ab=\left(a-b\right)^2\)

Do \(\left(a-b\right)^2\ge0\forall a,b\)

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

Dấu "=" xảy ra khi `a=b`

b. \(ab\le\frac{a^2+b^2}{2}\)

\(\Leftrightarrow2ab< a^2+b^2\)

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

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

Dấu "=" xảy ra khi `a-b=0<=>a=b`

c. Ta có: \(a\left(a+2\right)-\left(a+1\right)^2=a^2+2a-a^2-2a-1=-1\)

Do \(-1< 0\) (Luôn đúng)

\(\Rightarrow a^2+2a-a^2-2a-1< 0\)

\(\Rightarrow a^2+2a< a^2+2a+1\)

\(\Rightarrow a\left(a+2\right)< \left(a+1\right)^2\)

d. Ta có: \(\left(a-b+c\right)^2\ge0\forall a,b,c\)

\(\Rightarrow a^2+b^2+c^2-2ab+2ac-2bc\ge0\forall a,b,c\)

\(\Rightarrow a^2+b^2+c^2+\left(2ab-4ab\right)+\left(4ac-2ac\right)-2bc\ge0\forall a,b,c\)

\(\Rightarrow a^2+b^2+c^2+2ab-2ac-\left(4ab-4ac\right)\ge0\forall a,b,c\)

\(\Rightarrow\left(a+b-c\right)^2-\left(4ab-4ac\right)\ge0\forall a,b,c\)

\(\Rightarrow\left(a+b-c\right)^2\ge4ab-4ac\ge a,b,c\)

\(\Rightarrow\left(a+b-c\right)^2\ge4a\left(b-c\right)\)

e. Ta có: \(\hept{\begin{cases}\left(m-1\right)^2\ge0\forall m\\\left(n-1\right)^2\ge0\forall n\end{cases}}\)

\(\Rightarrow\left(m-1\right)^2+\left(n-1\right)^2\ge0\)

\(\Rightarrow m^2-2m+1+n^2-2n+1\ge0\)

\(\Rightarrow m^2+n^2+2\ge2n+2m\)

\(\Rightarrow m^2+n^2+2\ge2\left(n+m\right)\)

Dấu "=" xảy ra khi `m=n=1`

f. \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)

\(\Leftrightarrow1+\frac{b}{a}+1+\frac{a}{b}\ge4\)

\(\Leftrightarrow2+\frac{b}{a}+\frac{a}{b}\ge4\)

Áp dụng BĐT Cô-si cho hai số dương `a` và `b`, ta có:

\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}\)

\(\frac{a}{b}+\frac{b}{a}\ge2\)

Vậy \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\) khi \(a=b\)

g. Áp dụng BĐT Svac-xơ, ta có: 

\(\frac{1}{x}+\frac{4}{y}=\frac{1^2}{x}+\frac{2^2}{y}\ge\frac{\left(1+2\right)^2}{x+y}\)

Dấu "=" xảy ra khi \(\frac{1}{x}=\frac{4}{y}\Leftrightarrow4x=y\Leftrightarrow x=\frac{y}{4}\)

ĐKXĐ:  \(x\ne\pm1\)