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.

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)

\(\frac{yz+xz+xy}{xyz}=0\)

yz + xz + xy = 0

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2\times\left(xy+xz+yz\right)=x^2+y^2+z^2+2\times0=x^2+y^2+z^2\left(\text{đ}pcm\right)\)

a) Từ giả thiết suy ra: xy + yz + zx = 0

Do đó:

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=x^2+y^2+z^2\)

b) Đặt \(\frac{1}{a-b}=x\); \(\frac{1}{b-c}=y\); \(\frac{1}{c-a}=z\)

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a-b+b-c+c-a=0\)

Theo câu a ta có: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)

Suy ra điều phải chứng minh

a)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{xy+yz+xz}{xyz}=0\)

\(\Rightarrow xy+yz+xz=0\)

\(x^2+y^2+z^2=\left(x+y+z\right)^2\)

\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2+2xy+2yz+2xz\)

\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

Do \(xy+yz+xz=0\)

\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2\) ( đpcm )

b)

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

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

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

\(\Rightarrow2\left(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)}\right)=0\)

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

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

\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)\left[\left(a-b\right)+\left(b-c\right)+\left(c-a\right)\right]}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)

\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)\left[\left(-a+a\right)+\left(-b+b\right)+\left(-c+c\right)\right]}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)

\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right).0}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)

\(\Rightarrow0=0\) ( đpcm )

đặt x=a-b;y=b-c;z=c-a

ta có x+y+z=0

nên ta có ĐPCM 

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

<=> \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)

<=> \(2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=0\)

<=> \(\frac{z}{xyz}+\frac{y}{xyz}+\frac{x}{xyz}=0\)

<=> \(\frac{x+y+z}{xyz}=0\) (luôn đúng )

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)

\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)

\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)

\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)

Dấu "=" xảy ra khi x=y=z

Đặt: \(\hept{\begin{cases}\frac{1-a}{1+a}=x\\\frac{1-b}{1+b}=y\\\frac{1-c}{1+c}=z\end{cases}}\)

\(\Rightarrow-1< x,y,z< 1\)và \(\hept{\begin{cases}\frac{1-x}{1+x}=a\\\frac{1-y}{1+y}=b\\\frac{1-z}{1+z}=c\end{cases}}\)

Theo đề bài ta có: \(abc=1\Rightarrow\left(1-x\right)\left(1-y\right)\left(1-z\right)=\left(1+x\right)\left(1+y\right)\left(1+z\right)\)

\(\Rightarrow x+y+z+xyz=0\)

Mặt khác ta có: \(\frac{4a}{\left(a+1\right)^2}=1-x^2;\frac{2}{a+1}=1+x\)

Và: \(\frac{4b}{\left(b+1\right)^2}=1-y^2;\frac{2}{b+1}=1+y\)

Và: \(\frac{4c}{\left(c+1\right)^2}=1-z^2;\frac{2}{c+1}=1+z\)

Nên: \(\frac{4a}{\left(a+1\right)^2}+\frac{4b}{\left(b+1\right)^2}+\frac{4c}{\left(c+1\right)^2}\le1+2.\frac{2}{a+1}.\frac{2}{b+1}.\frac{2}{c+1}\)

\(\Leftrightarrow x^2+y^2+z^2+\left(xy+yz+zx\right)+2\left(x+y+z+xyz\right)\ge0\)

\(\Leftrightarrow\left(x+y+z\right)^2\ge0\)

Đây là BĐT luôn đúng nên ta có đpcm.

ミ★ᗪเệų ℌųуềй (ßăйǥ ßăйǥ ²к⁶)★彡 Giải ghê quá, t chẳng hiểu gì.

Đặt \(\left(a;b;c\right)=\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\)

BĐT \(\Leftrightarrow \sum\limits_{cyc} \frac{xy}{(x+y)^2} \leq \frac{1}{4}+\frac{4xyz}{(x+y)(y+z)(z+x)}\)

Ta có: \(VP-VT=\frac{4\left(x-y\right)^2\left(y-z\right)^2\left(z-x\right)^2}{4\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\ge0\)

BĐT hiển nhiên đúng.

Ôi trời, dòng 3 gõ latex mà olm không hiện à?

BĐT \(\Leftrightarrow\Sigma_{cyc}\frac{xy}{\left(x+y\right)^2}-\frac{1}{4}\le\frac{4xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

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

Cộng theo vế :

\(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

b/ \(\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{4}{a+2b+c}\)

\(\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{b+2c+a}\)

\(\frac{1}{c+a}+\frac{1}{a+b}\ge\frac{4}{c+b+2a}\)

Cộng theo vế :

\(2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge4\left(\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\right)\)

\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge2\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\)