Question

Bài 1. CM các BĐT sau

1. \(\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq ab + bc + ca\)

2. \(a, b, c, d, e > 0 \Rightarrow a(b+c+d+e) \geq 4\sqrt[4]{bcde}\)

3. \(\frac{1}{1+a^2} + \frac{1}{1+b^2} \leq \frac{1}{1+ab}\), với \(a, b > 1\)

Bài 2. Cho a, b, c, > 0, ab + bc + ca = 1. CM:

1. \(a^4 + b^4 + c^4 \geq abc\)

2. \(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{1}{2}\)

3. \(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \geq \frac{3}{2}\)

Bài 3. Cho a, b, c > 0, CM các BĐT:

1. \((\sqrt{a} + \sqrt{b})^2 \geq 2(\sqrt{a} + \sqrt{b})\sqrt{ab}\)

2. \((1+a+b)(1+b+c) \geq 9abc\)

Answer 1

Bài 2:

1) Ta có: `a^4+b^4+c^4>=a^2b^2+b^2c^2+c^2a^2`

Ta cần chứng minh: `a^2b^2+b^2c^2+c^2a^2>=abc` 

Theo BĐT Cauchy ta có:

`a^2b^2+b^2c^2>=2ab^2c`

`a^2b^2+c^2a^2>=2a^2bc`

`b^2c^2+c^2a^2>=2abc^2`

`=>2(a^2b^2+b^2c^2+c^2a^2)>=2(ab^2c+a^2bc+abc^2)`

`<=>a^2b^2+b^2c^2+c^2a^2>=abc(a+b+c)=abc` (đpcm) 

Answer 2

2) `a^2/(b+c)+b^2/(c+a)+c^2/(a+b)`

Áp dụng BĐT Cauchy Schwarz ta có:

`a^2/(b+c)+b^2/(c+a)+c^2/(a+b)>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2=1/2` (ĐPCM) 

3) Ta có bđt quen thuộc sau: `1/x+1/y+1/z>=9/(x+y+z)`

Áp dụng ta có: `2/(a+b)+2/(b+c)+2/(c+a)>=2*9/(a+b+b+c+c+a)=18/(2(a+b+c))=9` (ĐPCM)