Ta chứng minh BĐT sau với các số dương:

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Áp dụng:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)

Cộng vế với vế:

\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)

b.

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)

\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)

Cộng vế với vế (1); (2) và (3):

\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

a) Đặt: \(b+c=x;c+a=y;a+b=z\)

Có: \(x+y-z=b+c+c+a-a-b=2c\)

=> \(c=\frac{x+y-z}{2}\)

Tương tự ta cũng có:

\(a=\frac{y+z-x}{2};b=\frac{x+z-y}{2}\)

Có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

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

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

\(=\frac{1}{2}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)-3\right]\) (1)

Áp dụng bđt cô si ta có:

\(\frac{y}{x}+\frac{x}{y}\ge2;\frac{z}{x}+\frac{x}{z}\ge2;\frac{z}{y}+\frac{y}{z}\ge2\)

=> \(\left(1\right)\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)

Vậy \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)

b) Có: \(\frac{a^2}{b+c}+\frac{b+c}{4}=\frac{\left(2a\right)^2+\left(b+c\right)^2}{4\left(b+c\right)}\) (1)

VÌ: \(\left[2a-\left(b+c\right)\right]^2\ge0\)

=> \(\left(2a\right)^2+\left(b+c\right)^2\ge4a\left(b+c\right)\)

=> \(\left(1\right)\ge\frac{4a\left(b+c\right)}{4\left(b+c\right)}=a\)

Hay: \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge a\Rightarrow\frac{a^2}{b+c}\ge a-\frac{b+c}{4}\) (2)

Tương tự ta cũng có: \(\frac{b^2}{c+a}\ge b-\frac{c+a}{4}\) (3)

\(\frac{c^2}{a+b}\ge c-\frac{a+b}{4}\) (4)

Cộng vế với vế (2);(3);(4) ta có:

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

xin phép làm lại :3

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

\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(\ge\frac{1}{2}\cdot3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot\frac{3}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}-3=\frac{3}{2}\)( đpcm )

Dấu "=" xảy ra <=> a=b=c

Lời giải:
Ta có:

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

\(\Leftrightarrow \left(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}\right)+\left(\frac{b^2}{a^2+c^2}-\frac{b}{a+c}\right)+\left(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}\right)\geq 0\)

\(\Leftrightarrow \frac{ab(a-b)+ac(a-c)}{(b^2+c^2)(b+c)}+\frac{ba(b-a)+bc(b-c)}{(a^2+c^2)(a+c)}+\frac{ca(c-a)+cb(c-b)}{(a^2+b^2)(a+b)}\geq 0\)

\(\Leftrightarrow ab(a-b)\left(\frac{1}{(b^2+c^2)(b+c)}-\frac{1}{(a^2+c^2)(a+c)}\right)+bc(b-c)\left(\frac{1}{(a^2+c^2)(a+c)}-\frac{1}{(a^2+b^2)(a+b)}\right)+ca(c-a)\left(\frac{1}{(a^2+b^2)(a+b)}-\frac{1}{(b^2+c^2)(b+c)}\right)\geq 0\)

\(\Leftrightarrow ab(a-b).\frac{(a-b)(a^2+b^2+c^2+ab+bc+ac)}{(b^2+c^2)(b+c)(a^2+c^2)(a+c)}+bc(b-c).\frac{(b-c)(a^2+b^2+c^2+ab+bc+ac)}{(a^2+c^2)(a+c)(a^2+b^2)(a+b)}+ca(c-a).\frac{(c-a)(a^2+b^2+c^2+ab+bc+ac)}{(a^2+b^2)(a+b)(b^2+c^2)(b+c)}\geq 0\)

\(\Leftrightarrow (a^2+b^2+c^2+ab+bc+ac)\left[\frac{(a-b)^2}{(b^2+c^2)(b+c)(a^2+c^2)(a+c)}+\frac{(b-c)^2}{(a^2+c^2)(a+c)(a^2+b^2)(a+b)}+\frac{(c-a)^2}{(a^2+b^2)(a+b)(b^2+c^2)(b+c)}\right]\geq 0\)

(luôn đúng)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

Câu này quá khó .Thần đồng chắc mới giải được.

\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)

Tương tự: \(\frac{b^2}{a+c}+\frac{a+c}{4}\ge b\) ; \(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)

Cộng vế với vế:

\(VT+\frac{a+b+c}{2}\ge a+b+c\Rightarrow VT\ge\frac{a+b+c}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)

Do vai trò của a,b,c là như nhau nên ta cò thể giả sử: \(a\ge b\ge c>0\)

Ta có:\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}\)

CMTT: \(\frac{b^2}{c^2+a^2}-\frac{b}{c+a}=\frac{bc\left(b-c\right)-ab\left(a-b\right)}{\left(a+c\right)\left(a^2+c^2\right)}\)

             \(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{-bc\left(b-c\right)-ac\left(a-c\right)}{\left(a+b\right)\left(a^2+b^2\right)}\)

Đặt \(A=\left(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\right)-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)

\(\Rightarrow A=\left[\frac{ab\left(a-b\right)}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{ab\left(a-b\right)}{\left(a+c\right)\left(a^2+c^2\right)}\right]\)+ \(\left[\frac{ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{ac\left(a-c\right)}{\left(a+b\right)\left(a^2+b^2\right)}\right]\)+       \(\left[\frac{bc\left(b-c\right)}{\left(a+c\right)\left(a^2+c^2\right)}-\frac{bc\left(b-c\right)}{\left(a+b\right)\left(a^2+b^2\right)}\right]\)

\(\Rightarrow A=ab\left(a-b\right)\left[\frac{1}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+c\right)\left(a^2+c^2\right)}\right]^{\left(1\right)}\)+ ... 

Do \(a\ge b\ge c>0\Rightarrow\left(1\right)>0.\)

CMTT \(\Rightarrow A>0.\Rightarrowđpcm\)

(Mình làm hơi tắt, mong bạn thông cảm. Cho 1 k nha.)

​​

Tại sao (1) lại >0 hả bạn? Với lại đề mình cho đâu có đk a>=b>=c>0 đâu!

Đề bài cho a,b,c>0.Do chúng có vai trò như nhau nên mình giả sử như trên.Do \(a\ge b>0\Rightarrow ab;\left(a-b\right)>0\)

Lại có \(a\ge b\ge c>0\Rightarrow a+c>b+c;a^2+c^2>b^2+c^2\)

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

\(\Rightarrow\frac{1}{\left(a+c\right)\left(a^2+c^2\right)}< \frac{1}{\left(b+c\right)\left(b^2+c^2\right)}\)

\(\Rightarrow\frac{1}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+c\right)\left(a^2+c^2\right)}>0\)

\(\Rightarrow\left(1\right)>0\)

2) Theo nguyên lí Dirichlet, trong ba số \(a^2-1;b^2-1;c^2-1\) có ít nhất hai số nằm cùng phía với 1.

Giả sử đó là a² - 1 và b² - 1. Khi đó \(\left(a^2-1\right)\left(b^2-1\right)\ge0\Leftrightarrow a^2b^2-a^2-b^2+1\ge0\)

\(\Rightarrow a^2b^2+3a^2+3b^2+9\ge4a^2+4b^2+8\)

\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\ge4\left(a^2+b^2+2\right)\)

\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\) (2)

Mà \(4\left[\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\right]\ge4\left(a+b+c+1\right)^2\) (3)(Áp dụng Bunhicopxki và cái ngoặc vuông)

Từ (2) và (3) ta có đpcm.

Sai thì chịu

Xí quên bài 2 b:v

b) Không mất tính tổng quát, giả sử \(\left(a^2-\frac{1}{4}\right)\left(b^2-\frac{1}{4}\right)\ge0\)

Suy ra \(a^2b^2-\frac{1}{4}a^2-\frac{1}{4}b^2+\frac{1}{16}\ge0\)

\(\Rightarrow a^2b^2+a^2+b^2+1\ge\frac{5}{4}a^2+\frac{5}{4}b^2+\frac{15}{16}\)

Hay \(\left(a^2+1\right)\left(b^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{3}{4}\right)\)

Suy ra \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+\frac{1}{4}+c^2+\frac{1}{2}\right)\)

\(\ge\frac{5}{4}\left(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c+\frac{1}{2}\right)^2=\frac{5}{16}\left(a+b+c+1\right)^2\) (Bunhiacopxki) (đpcm)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)

Cách nữa cho bài 2:

2a) Ta có: \(4\left(a^2+1+2\right)\left(1+1+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2\)

Hay \(4\left(a^2+3\right)\left(2+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2=VP\)

Như vậy ta quy bài toán về chứng minh: \(\left(b^2+3\right)\left(c^2+3\right)\ge4\left(2+\frac{\left(b+c\right)^2}{2}\right)\)

\(\Leftrightarrow b^2c^2+b^2+c^2+1\ge4bc\Leftrightarrow\left(bc-1\right)^2+\left(b-c\right)^2\ge0\)(đúng)

Đẳng thức xảy ra khi a = b = c = 1

b) Áp dụng BĐT Bunhiacopxki:\(\left(a^2+\frac{1}{4}+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+b^2+c^2+\frac{1}{2}\right)\ge\frac{1}{4}\left(a+b+c+1\right)^2\)

\(\Rightarrow\frac{5}{4}\left(a^2+1\right)\left(b^2+c^2+\frac{3}{4}\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)

Từ đó ta có thể quy bài toán về chứng minh: \(\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(b^2+c^2+\frac{3}{4}\right)\)

...

Bài 3:Sửa đề a, b, c >0

Có:  \(\frac{a^3}{b^2}+\frac{a^3}{b^2}+b\ge3\sqrt[3]{\frac{a^6}{b^3}}=\frac{3a^2}{b}\)

Tương tự: \(\frac{2b^3}{c^2}+c\ge\frac{3b^2}{c};\frac{2c^3}{a^2}+a\ge\frac{3c^2}{a}\)

Cộng theo vế 3 BĐT trên: \(2\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)+a+b+c\ge3\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)

\(=2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)

\(\ge2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+a+b+c\)

Từ đó ta có đpcm.