Không mất tính tổng quát, chuẩn hóa a + b + c = 1

Khi đó, ta cần chứng minh: \(\frac{\left(a+1\right)^2}{2a^2+\left(1-a\right)^2}+\frac{\left(b+1\right)^2}{2b^2+\left(1-b\right)^2}+\frac{\left(c+1\right)^2}{2c^2+\left(1-c\right)^2}\le8\)

Xét bất đẳng thức phụ: \(\frac{\left(x+1\right)^2}{2x^2+\left(1-x\right)^2}\le4x+\frac{4}{3}\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{\left(3x-1\right)^2\left(4x+1\right)}{2x^2+\left(1-x\right)^2}\ge0\)*đúng*

Áp dụng, ta được: \(\frac{\left(a+1\right)^2}{2a^2+\left(1-a\right)^2}+\frac{\left(b+1\right)^2}{2b^2+\left(1-b\right)^2}+\frac{\left(c+1\right)^2}{2c^2+\left(1-c\right)^2}\)\(\le4\left(a+b+c\right)+4=4.1+4=8\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi a = b = c

Chuẩn hóa ta có : \(a+b+c=3\)

=> \(\frac{\left(2a+b+c\right)^2}{2a^2+\left(b+c\right)^2}=\frac{\left(a+3\right)^2}{2a^2+\left(3-a\right)^2}=\frac{a^2+6a+9}{3\left(a^2-2a+3\right)}\)

Xét\(\frac{a^2+6a+9}{3\left(a^2-2a+3\right)}\le\frac{4}{3}a+\frac{4}{3}\)

<=> \(a^2+6a+9\le4\left(a+1\right)\left(a^2-2a+3\right)\)

<=> \(4a^3-5a^2-2a+3\ge0\)

<=> \(\left(a-1\right)^2\left(4a+3\right)\ge0\)luôn đúng

Khi đó 

\(VT\le\frac{4}{3}\left(a+b+c\right)+4=\frac{4}{3}.3+4=8\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c

bài lớp 10 em chưa hok nha anh

Ta có 

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

\(\Rightarrow VT\ge\frac{3}{2}\left[\left(a+b\right)^2.c^2+4+2\left(a+b\right)^2+2c^2\right]\)

           \(\ge\frac{3}{2}\left[4\left(a+b\right)c+2\left(a+b\right)^2+2c^2\right]=VP\)

=> ĐPCM

Dấu "=" xảy ra khi 

\(a=b=c=\frac{\pm1}{\sqrt{2}}\)

Đề PBC 2015-2016 nè

t nói trước đây là bài làm rất xàm nên đừng tin nhé,spam đấy!

Không mất tính tổng quát giả sử \(c\ge0\)

\(a=c+x+y;b=c+y;c=c\)

Ta cần chứng minh \(A=f\left(x;y;c\right)=\left[\left(c+x+y\right)^2+\left(c+y\right)^2+c^2\right]\left[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{\left(x+y\right)^2}\right]\ge\frac{9}{2}\)

\(\ge\frac{\left(3c+x+y\right)}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{\left(x+y\right)^2}\right)=T\left(x;y;c\right)\)

Xét hiệu \(T\left(x;y;c\right)-T\left(x;y;0\right)=c\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{\left(x+y\right)^2}\right)\ge0\)

Nên \(T\left(x;y;c\right)\ge T\left(x;y;0\right)=\frac{1}{3}\left(x+y\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{\left(x+y\right)^2}\right)\)

Cần chứng minh \(\frac{1}{3}\left(x+y\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{\left(x+y\right)^2}\right)\ge\frac{9}{2}\)

...

đây là toán mà bn

Đặt: b + c - a = x; a + b - c = y; a + c - b = z

khi đó: x + y + z = a + b + c 

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

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

Ta cần chứng minh: 

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

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

Hay mình cần chứng minh: 

\(\frac{y+z}{2}.\frac{y-z}{2}.x^2+\frac{z+x}{2}.\frac{z-x}{2}.y^2+\frac{x+y}{2}.\frac{x-y}{2}.z^2=0\)

<=> \(\left(y^2-z^2\right)x^2+\left(z^2-x^2\right)y^2+\left(x^2-y^2\right)z^2=0\)

<=> \(0=0\)luôn đúng

Vậy (1) đúng

tìm lag mắt mới ra Xem câu hỏi

Tương tự, ta được:

\(\left(2-y\right)\left(2-z\right)>=\dfrac{\left(x+1\right)^2}{4}\)

và \(\left(2-z\right)\left(2-x\right)>=\left(\dfrac{y+1}{2}\right)^2\)

=>8(2-x)(2-y)(2-z)>=(x+1)(y+1)(z+1)

(x+yz)(y+zx)<=(x+y+yz+xz)^2/4=(x+y)^2*(z+1)^2/4<=(x^2+y^2)(z+1)^2/4

Tương tự, ta cũng co:

\(\left(y+xz\right)\left(z+y\right)< =\dfrac{\left(y^2+z^2\right)\left(x+1\right)^2}{2}\)

và \(\left(z+xy\right)\left(x+yz\right)< =\dfrac{\left(z^2+x^2\right)\left(y+1\right)^2}{2}\)

Do đó, ta được:

\(\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)< =\left(x+1\right)\left(y+1\right)\left(z+1\right)\)

=>ĐPCM

1. Ta có: \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)\)

\(=2a.2b=4ab\)

=> đpcm

2. Ta có: \(\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2\)

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

=> đpcm

3. Ta có:\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab\)

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

=> đpcm

4. Ta có: \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)

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

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

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

\(\Leftrightarrow a^2+b^2-a^2-b^2+2ab+2ab=4ab\)

\(\Leftrightarrow4ab=4ab\Leftrightarrow4ab-4ab=0\Leftrightarrow0=0\)(đpcm)

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

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

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

\(\Leftrightarrow2\left(a^2+b^2\right)=2\left(a^2+b^2\right)\Leftrightarrow2\left(a^2+b^2\right)-2\left(a^2+b^2\right)=0\Leftrightarrow0=0\)(đpcm)

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

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

\(\Leftrightarrow a^2+b^2-2ab=a^2+b^2-2ab\)

\(\Leftrightarrow\left(a-b\right)^2=\left(a-b\right)^2\Leftrightarrow\left(a-b\right)^2-\left(a-b\right)^2=0\Leftrightarrow0=0\)(đpcm)

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

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

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

\(\Leftrightarrow a^2+b^2+2ab=\left(a+b\right)^2\Leftrightarrow\left(a+b\right)^2=\left(a+b\right)^2\)

\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)^2=0\Leftrightarrow0=0\)(đpcm)

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

\(=2b.2a=4ab\)( đpcm )

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

\(=2\left(a^2+b^2\right)\)( đpcm )

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

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

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

\(=a^2+2ab+b^2=\left(a+b\right)^2\)( đpcm )

đặt \(\hept{\begin{cases}a+b=x\\b+c=y\\c+a=z\end{cases}}\)

cậu tính A theo x,y,x rồi chứng minh 

\(B=\frac{x}{z-y}.\frac{y}{x-z}+\frac{y}{x-z}.\frac{z}{y-x}+\frac{z}{y-x}.\frac{x}{z-y}=-1\)

thì ta có A+2B>=0   -->A>=-2B=2

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

Subtract 2 from both sides:

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

Refine:

\(\frac{\left(a+b\right)^2}{a-b}+\frac{\left(b+c\right)^2}{b-c}+\frac{\left(c+a\right)^2}{c-a}\ge0\)

Simplyfy : \(\frac{\left(a+b\right)^2}{\left(a-b\right)}+\frac{\left(b+c\right)^2}{b-c}+\frac{\left(c+a\right)^2}{c-a}:\)     \(\frac{4a^2bc-4a^2c^2-4a^2b^2+2a^2b-2a^2c+4ab^2c+4abc^2+2ac^2-2ab^2-4b^2c^2+2b^2c-2bc^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

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

Convert element to fraction: \(2=\frac{2}{1}\)

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

Find LCD for: \(\frac{\left(a+b\right)^2}{\left(a-b\right)}+\frac{\left(b+c\right)^2}{\left(b-c\right)}+\frac{\left(c+a\right)^2}{c-a}-\frac{2}{1}\):

Find the least common denominator 1   (a  - b) (b - c) (c- a) = (a  - b) (b - c) (c- a)(a  - b) (b - c) (c- a)

Sau đó vào đây để xem bài giải tiếp theo nhá! Lười đánh máy tiếp lắm!   Có gì mai mốt sử dụng phần mềm đó giải khỏi phải lên đây hỏi.

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