§1. Bất đẳng thức

m.n ơi giúp mk 1 hoặc 2 câu đc ko ạ mk cần gấp lắm mà mk ko bt cách lm

Em xin phép nhờ  quý thầy cô và các bạn giúp đỡ với ạ!

Đặt \(a\left(1-b\right)=x;b\left(1-c\right)=y;c\left(1-a\right)=x\)

\(\Rightarrow1-\left(a+b+c\right)+ab+bc+ca=1-a\left(1-b\right)-b\left(1-c\right)-c\left(1-a\right)=1-x-y-z\)

BĐT cần c/m trở thành:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{3}{1-x-y-z}\)

\(\Leftrightarrow\left(1-x-y-z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-3\ge0\)

\(\Leftrightarrow\dfrac{1-x-y-z}{x}+\dfrac{1-x-y-z}{y}+\dfrac{1-x-y-z}{z}-3\ge0\)

\(\Leftrightarrow\dfrac{1-y-z}{x}+\dfrac{1-z-x}{y}+\dfrac{1-x-y}{z}-6\ge0\) (1)

Lại có: \(1-y-z=1-b\left(1-c\right)-c\left(1-a\right)=1-b-c+bc+ca=\left(1-b\right)\left(1-c\right)+ca\)

Nên (1) tương đương:

\(\dfrac{\left(1-b\right)\left(1-c\right)+ca}{a\left(1-b\right)}+\dfrac{\left(1-a\right)\left(1-c\right)+ab}{b\left(1-c\right)}+\dfrac{\left(1-a\right)\left(1-b\right)+bc}{c\left(1-a\right)}-6\ge0\)

\(\Leftrightarrow\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\)

BĐT trên hiển nhiên đúng theo AM-GM do:

\(\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\sqrt[6]{\dfrac{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}}=6\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)

ráng chờ thầy nguyễn việt lâm  onl r nhờ nghen:>

\(\sqrt{\left(16+9\right)\left(16a^2b^2+9\right)}\ge\sqrt{\left(16ab+9\right)^2}=16ab+9\)

\(\Rightarrow\sqrt{16a^2b^2+9}\ge\dfrac{1}{5}\left(16ab+9\right)\)

\(\Rightarrow P\ge\dfrac{1}{5}\left(16ab+9\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{5}\left[16\left(a+b\right)+9\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]\)

\(P\ge\dfrac{1}{5}\left[32+9.\dfrac{4}{a+b}\right]=\dfrac{1}{5}\left[32+\dfrac{9.4}{2}\right]=10\)

\(P_{min}=10\) khi \(a=b=1\)

\(VT=\sqrt{\left(a+\dfrac{5b}{2}\right)^2+\dfrac{15b^2}{4}}+\sqrt{\left(b+\dfrac{5c}{2}\right)^2+\dfrac{15c^2}{4}}+\sqrt{\left(c+\dfrac{5a}{2}\right)^2+\dfrac{15a^2}{4}}\)

\(\Rightarrow VT\ge\sqrt{\left(a+\dfrac{5b}{2}+b+\dfrac{5c}{2}+c+\dfrac{5a}{2}\right)^2+\dfrac{15}{4}\left(a+b+c\right)^2}\)

\(\Rightarrow VT\ge\sqrt{\dfrac{49}{4}\left(a+b+c\right)^2+\dfrac{15}{4}\left(a+b+c\right)^2}=4\left(a+b+c\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

Áp dụng BĐT Bunhiacopxki (cho tất cả các bài):

1.

\(\left(3x+4y\right)^2\le\left(3^2+4^2\right)\left(x^2+y^2\right)=25\)

\(\Rightarrow\left|3x+4y\right|\le5\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{3}{5};\dfrac{4}{5}\right)\)

2.

\(\left(x+2y\right)^2=\left(1.x+\sqrt{2}.\sqrt{2y}\right)^2\le\left(1+2\right)\left(x^2+2y^2\right)=3\)

\(\Rightarrow\left|x+2y\right|\le\sqrt{3}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{\sqrt{3}};\dfrac{1}{\sqrt{3}}\right)\)

4.

a.

Áp dụng Bunhiacopxki:

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

\(\Rightarrow2\left(a+b+c\right)\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)

3.

Từ giả thiết:

\(x^2+y^2+z^2-2x-4y+2z=3\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2+\left(z+1\right)^2=9\)

Ta có:

\(\left|x+2y+2z-20\right|=\left|x-1+2\left(y-2\right)+2\left(z+1\right)-17\right|\)

\(\Rightarrow\left|x+2y+2z-20\right|\le\left|x-1+2\left(y-2\right)+2\left(z+1\right)\right|+17\)

Lại có:

\(\left[\left(x-1\right)+2\left(y-2\right)+2\left(z+1\right)\right]^2\le\left(1+4+4\right)\left[\left(x-1\right)^2+\left(y-2\right)^2+\left(z+1\right)^2\right]=81\)

\(\Rightarrow\left|x-1+2\left(y-2\right)+2\left(z+1\right)\right|\le9\)

\(\Rightarrow\left|x+2y+2z-20\right|\le9+17=26\)

\(\Rightarrow A_{max}=26\) khi \(\left(x;y;z\right)=\left(0;0;-3\right)\)

Đặt \(P=\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{a+d}+\dfrac{d}{a+b}\)

\(P=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{ac+cd}+\dfrac{d^2}{ad+bd}\)

\(P\ge\dfrac{\left(a+b+c+d\right)^2}{ab+2ac+bc+2bd+cd+ad}=\dfrac{\left(a+c\right)^2+\left(b+d\right)^2+2\left(a+c\right)\left(b+d\right)}{2ac+2bd+ab+bc+cd+ad}\)

\(P\ge\dfrac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=d\)

Bài toán cơ bản:

\(abc=1\Rightarrow\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\) 

Bunhiacopxki:

\(\left(a+b+c\right)\left(\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right)\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\right)^2=1\)

\(\Rightarrow\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\ge\dfrac{1}{a+b+c}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Cách 2:

Do \(abc=1\), đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

Ta có \(\dfrac{a}{\left(ab+a+1\right)^2}=\dfrac{\dfrac{x}{y}}{\left(\dfrac{x}{z}+\dfrac{x}{y}+1\right)^2}=\dfrac{\dfrac{x}{y}.y^2z^2}{\left(xy+yz+zx\right)^2}=\dfrac{xyz^2}{\left(xy+yz+zx\right)^2}\)...

Từ đó, BĐT cần chứng minh trở thành:

\(\dfrac{xyz^2}{\left(xy+yz+zx\right)^2}+\dfrac{x^2yz}{\left(xy+yz+zx\right)^2}+\dfrac{xy^2z}{\left(xy+yz+zx\right)^2}\ge\dfrac{1}{\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}}\)

\(\Leftrightarrow xyz\left(x+y+z\right)\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\ge\left(xy+yz+zx\right)^2\)

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

Thật vậy, áp dụng BĐT Bunhiacopxki:

\(\left(z+x+y\right)\left(x^2z+y^2x+z^2y\right)\ge\left(\sqrt{zx^2z}+\sqrt{xy^2x}+\sqrt{yz^2y}\right)^2=\left(xy+yz+zx\right)^2\) (đpcm)