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

\(\frac{2\left(a+b-1\right)}{\left(a+b\right)^2-1}+2=\frac{2}{a+b+1}+2\ge\frac{2}{\sqrt{2\left(a^2+b^2\right)}+1}+2=\frac{2}{\sqrt{2}+1}+2=2\sqrt{2}\)

Dấu = xảy ra khi \(a=b=\frac{1}{\sqrt{2}}\)

Đặt \(a=\frac{x^2}{z},b=\frac{y^2}{z}\rightarrow x^4+y^4=z^2\) where x, y, z> 0

\(z\left(\frac{1}{x^2}+\frac{1}{y^2}\right)-\left(\frac{x}{y}-\frac{y}{x}\right)^2\ge2\sqrt{2}\)

\(\Leftrightarrow\sqrt{x^4+y^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\ge2\sqrt{2}+\left(\frac{x}{y}-\frac{y}{x}\right)^2\)

\(\Leftrightarrow\frac{2\left(3-2\sqrt{2}\right)\left(x^2-y^2\right)^2}{x^2y^2}\ge0\) *Đúng*

Cái này chỉ có tìm x thôi

Ta có \(x=2-\sqrt{3}\)

=>\(x^2-4x=\left(2-\sqrt{3}\right)^2-4\left(2-\sqrt{3}\right)=4-4\sqrt{3}+3-8+4\sqrt{3}=-1\)

Thay vào A ta được:

\(A=7\left(-1\right)^{100}+\left(-1\right)^{50}+2016=7+1+2016=2024\)

Vậy A=2024

Đặt \(y=\frac{2x+4}{\sqrt{1-x^2}}\) (ĐKXĐ: -1<x<1)

<=> \(y^2=\frac{2x^2+8x+8}{1-x^2}\)

<=>\(y^2-x^2y^2=2x^2+8x+8\)

<=>\(y^2\left(1-x^2\right)-2x^2-8x-8=0\)

Xét \(∆=0-4.\left(1-x^2\right)\left(-2x^2-8x-8\right)=-8x^4-16x^3-24x^2+16x+32\)

Mà ∆≥0

<=>\(x^4+2x^3+3x^2-2x-4\le0\)

<=>....

ko biết

\(D=\frac{1}{a^2+b^2-2a-2b+2}+\frac{1}{ab-a-b+1}+4\left(ab-a-b\right)\)

\(=\frac{1}{a^2+b^2-2a-2b+2}+\frac{1}{2ab-2a-2b+2}+\frac{1}{2\left(ab-a-b+1\right)}+4\left(ab-a-b\right)\)

\(\ge\frac{4}{a^2+b^2-4a-4b+2ab+4}+\frac{1}{2\left(ab-a-b+1\right)}+8\left(ab-a-b+1\right)-4\left(ab-a-b+1\right)-4\)

\(\ge\frac{4}{\left(a+b-2\right)^2}+2\sqrt{\frac{1}{2\left(ab-a-b+1\right)}.8\left(ab-a-b+1\right)}-4\left(ab-a-b+1\right)-4\)

\(\ge4+4-4\left(ab-a-b+1\right)-4\)

= 4 ( a + b ) - 4ab 

\(\ge\)4 ( a + b ) - (a + b )² - 4 + 4

=  - ( a + b - 2 )^2 + 4 

\(\ge\)3

Dấu "=" <=> a = b = 3/2

Đưa D về dạng: 

D = \(\frac{1}{\left(a-1\right)^2+\left(b-1\right)^2}+\frac{1}{\left(a-1\right)\left(b-1\right)}+4\left(a-1\right)\left(b-1\right)-4\)

\(\left(a-1\right)+\left(b-1\right)=a+b-2\le1\)

Đặt: a - 1 = x ; b - 1 = y => x + y \(\le\)1

=> \(D=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy-4\)

Tìm min D. Làm như này chắc nhanh hơn. Bạn thử xem nhé!

bạn không nên đưa những câu hỏi linh tinh,vớ  vẩn lên diễn đàn nha

Bài làm:

Ta có: \(C=\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\ge\frac{\left(1+2+3\right)^2}{x+y+z}\ge\frac{6^2}{1}=36\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=\frac{1}{13}\\y=\frac{4}{13}\\z=\frac{9}{13}\end{cases}}\)

Bài này là áp dụng bđt Cauchy-Schwaz nha bạn.

huyen