Chính bài của em:

Cho \(a,b,c\ge1\). CMR: \(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)+2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}... - Hoc24

Không mất tính tổng quát ta giả sử \(a\ge b\ge c\)

Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\a-c=z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}z\ge x\ge0\\z\ge y\ge0\end{matrix}\right.\)

Ta có:

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

\(\Leftrightarrow x^2+y^2+z^2+2xz+2yz-2xy=0\)

\(\Leftrightarrow z^2+2xz+2yz+\left(x-y\right)^2=0\)

Vì \(\Rightarrow\left\{{}\begin{matrix}z\ge x\ge0\\z\ge y\ge0\end{matrix}\right.\)

\(\Rightarrow z^2+2xz+2yz+\left(x-y\right)^2\ge0\)

Dấu = xảy ra khi \(x=y=z=0\)

Hay \(a=b=c\)

\(VT=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4ab-4bc-4ca\)

\(VP=\left[\left(a+b\right)-2c\right]^2+\left[\left(b+c\right)-2a\right]^2+\left[\left(c+a\right)-2b\right]^2\)

\(=\left(a+b\right)^2-4\left(a+b\right)c+4c^2+\left(b+c\right)^2-4\left(b+c\right)a+4a^2+\left(a+c\right)^2-4\left(a+c\right)b+4b^2\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4\left(a+b\right)c+4c^2-4\left(b+c\right)a+4a^2-4\left(a+c\right)b+4b^2\)

Nhìn vào thấy 2 vế có \(\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\) rút gọn luôn thì được

\(-4ab-4bc-4ca=-4\left(a+b\right)c+4c^2-4\left(b+c\right)a+4a^2-4\left(a+c\right)b+4b^2\)

\(\Rightarrow ab-\left(a+b\right)c+c^2+bc-\left(b+c\right)a+a^2+ac-\left(a+c\right)c+b^2=0\)

\(\Rightarrow ab-ac-bc+c^2+bc-ab-ac+a^2+ac-ab-bc+b^2=0\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

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

Xảy ra khi \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(a+c-2b\right)^2\)\(< =>\left(a+b-2c\right)^2-\left(a-b\right)^2+\left(b+c-2a\right)^2-\left(b-c\right)^2+\left(a+c-2b\right)^2-\left(c-a\right)^2=0\)

<=> \(\left(2b-2c\right)\left(2a-2c\right)+\left(2b-2a\right)\left(2c-2a\right)+\left(2c-2b\right)\left(2a-2b\right)=0\)

<=> \(\left(b-c\right)\left(a-c\right)+\left(b-a\right)\left(c-a\right)+\left(c-b\right)\left(a-b\right)=0\)

<=> \(ab-ac-bc+c^2+bc-ac-ab+a^2+ac-ab-bc+b^2=0\)

<=> \(a^2+b^2+c^2-ab-ac-bc=0\)

<=> \(2a^{^2}+2b^2+2c^2-2ab-2ac-2bc=0\)

<=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=> a = b= c ( đpcm )

\(a^3+b^3+c^3-3abc\)

\(=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-\left(3a^2b+3ab^2+3abc\right)\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)\(\left(đpcm\right)\)

\(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Biến đổi VT=\(3\left(ab+bc+ca\right)-abc\left(a+b+c\right)=3\left(ab+bc+ca\right)-\frac{\left(ab+bc+ca\right)^2-a^2b^2-b^2c^2-c^2a^2}{2}\)

\(\le3t-\frac{t^2}{2}+\frac{3}{2}=\frac{12-\left(t-3\right)^2}{2}\le6\)(t=ab+bc+ca)

(a^2b^2+b^2c^2+c^2a^2 nhỏ hơn hoặc bằng 3)

Giải:

Ta có:

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=4a^2+4b^2+4c^2-4ab-4bc-4ac\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Leftrightarrow\left(a^2+b^2+c^2-ab-bc-ca\right)=2\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Rightarrowđpcm\)

Ta có:

\(\left(a-b\right)^2\)+\(\left(b-c\right)^2\)+\(\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\Leftrightarrow-2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)

\(\Leftrightarrow a=b=c\left(đpcm\right)\)

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.

nhiều thế, đăng ít một thôi bạn

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)

e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)

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

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

\(=4\left(a+b\right)^2-2\left(\left(a+b\right)^2-c^2\right)-2\left(a+b\right)^2\)

\(=4\left(a+b\right)^2-2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2=2c^2\)

g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)

\(=\left(2a+2b\right)^2-2\left(a+b+c+d\right)\left(a+b-c-d\right)+\left(2a-2b\right)^2-2\left(a+c-b-d\right)\left(a+d-b-c\right)\)

\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)

\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)

\(=2\left(\left(2a\right)^2-2\left(a+b\right)\left(a-b\right)\right)+2\left(\left(2c\right)^2-2\left(c+d\right)\left(c-d\right)\right)\)

\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)

\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)

bn đăng nhiều quá nên mk làm câu nào hay câu đó nha

mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .