Bài 4: Áp dụng bất đẳng thức AM - GM, ta có: \(P=\text{​​}\Sigma_{cyc}a\sqrt{b^3+1}=\Sigma_{cyc}a\sqrt{\left(b+1\right)\left(b^2-b+1\right)}\le\Sigma_{cyc}a.\frac{\left(b+1\right)+\left(b^2-b+1\right)}{2}=\Sigma_{cyc}\frac{ab^2+2a}{2}=\frac{1}{2}\left(ab^2+bc^2+ca^2\right)+3\)Giả sử b là số nằm giữa a và c thì \(\left(b-a\right)\left(b-c\right)\le0\Rightarrow b^2+ac\le ab+bc\)\(\Leftrightarrow ab^2+bc^2+ca^2\le a^2b+abc+bc^2\le a^2b+2abc+bc^2=b\left(a+c\right)^2=b\left(3-b\right)^2\)

Ta sẽ chứng minh: \(b\left(3-b\right)^2\le4\)(*)

Thật vậy: (*)\(\Leftrightarrow\left(b-4\right)\left(b-1\right)^2\le0\)(đúng với mọi \(b\in[0;3]\))

Từ đó suy ra \(\frac{1}{2}\left(ab^2+bc^2+ca^2\right)+3\le\frac{1}{2}.4+3=5\)

Đẳng thức xảy ra khi a = 2; b = 1; c = 0 và các hoán vị

Bài 1: Đặt \(a=xc,b=yc\left(x,y>0\right)\)thì điều kiện giả thiết trở thành \(\left(x+1\right)\left(y+1\right)=4\)

Khi đó  \(P=\frac{x}{y+3}+\frac{y}{x+3}+\frac{xy}{x+y}=\frac{x^2+y^2+3\left(x+y\right)}{xy+3\left(x+y\right)+9}+\frac{xy}{x+y}\)\(=\frac{\left(x+y\right)^2+3\left(x+y\right)-2xy}{xy+3\left(x+y\right)+9}+\frac{xy}{x+y}\)

Có: \(\left(x+1\right)\left(y+1\right)=4\Rightarrow xy=3-\left(x+y\right)\)

Đặt \(t=x+y\left(0< t< 3\right)\Rightarrow xy=3-t\le\frac{\left(x+y\right)^2}{4}=\frac{t^2}{4}\Rightarrow t\ge2\)(do t > 0)

Lúc đó \(P=\frac{t^2+3t-2\left(3-t\right)}{3-t+3t+9}+\frac{3-t}{t}=\frac{t}{2}+\frac{3}{t}-\frac{3}{2}\ge2\sqrt{\frac{t}{2}.\frac{3}{t}}-\frac{3}{2}=\sqrt{6}-\frac{3}{2}\)với \(2\le t< 3\)

Vậy \(MinP=\sqrt{6}-\frac{3}{2}\)đạt được khi \(t=\sqrt{6}\)hay (x; y) là nghiệm của hệ \(\hept{\begin{cases}x+y=\sqrt{6}\\xy=3-\sqrt{6}\end{cases}}\)

Ta lại có \(P=\frac{t^2-3t+6}{2t}=\frac{\left(t-2\right)\left(t-3\right)}{2t}+1\le1\)(do \(2\le t< 3\))

Vậy \(MaxP=1\)đạt được khi t = 2 hay x = y = 1

3. Áp dụng cô si ta có 

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

Lại có:

 \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=1\)

⇒ P ≥ \(2020.1+1=2021\)

Vậy Pmin = 2021 khi và chỉ khi a = b = c =1/3

vào thống kê để xem hình ảnh

\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)

\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)

Áp dụng BĐT Cosi ta có:

\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)

Từ (1)(2)(3) ta có:

\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)

Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)

Dấu "=" xảy ra <=> a=b=c=1

CHÚC BAN HỌC GIỎI

đây\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(\frac{b\left(2a-b\right)}{a\left(b+c\right)}+\frac{c\left(2b-c\right)}{b\left(c+a\right)}+\frac{a\left(2c-a\right)}{c\left(a+b\right)}\le\frac{3}{2}\)

\(\Leftrightarrow\left[2-\frac{b\left(2a-b\right)}{a\left(b+c\right)}\right]+\left[2-\frac{c\left(2b-c\right)}{b\left(c+a\right)}\right]+\left[2-\frac{a\left(2c-a\right)}{c\left(a+b\right)}\right]\ge\frac{9}{2}\)

\(\Leftrightarrow\frac{b^2+2ca}{a\left(b+c\right)}+\frac{c^2+2ab}{b\left(c+a\right)}+\frac{a^2+2bc}{c\left(a+b\right)}\ge\frac{9}{2}\)

Áp dụng BĐT Schwarz, ta có :

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

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

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

Cộng ( 1 ) với ( 2 ), ta được :

\(\frac{b^2+2ca}{a\left(b+c\right)}+\frac{c^2+2ab}{b\left(c+a\right)}+\frac{a^2+2bc}{c\left(a+b\right)}\)

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

\(\ge\left(a+b+c\right)^2\left(\frac{\left(1+2\right)^2}{2\left(ab+bc+ac\right)+2\left(a^2+b^2+c^2+ab+bc+ac\right)}\right)=\frac{9}{2}\)

không biết cách này ổn không 

Ta có : \(\frac{b\left(2a-b\right)}{a\left(b+c\right)}=\frac{2-\frac{b}{a}}{\frac{c}{b}+1}\) ; tương tự :...

đặt \(\frac{a}{c}=x;\frac{b}{a}=y;\frac{c}{b}=z\Rightarrow xyz=1\)

\(\Sigma\frac{2-y}{z+1}\le\frac{3}{2}\)          

\(\Leftrightarrow2\Sigma xy^2+2\Sigma x^2+\Sigma xy\ge3\Sigma x+6\)( quy đồng khử mẫu )

\(\Leftrightarrow\Sigma\frac{x}{y}\ge\Sigma x\)( xyz = 1 )           ( luôn đúng )

\(\Rightarrowđpcm\)

1.\(\left(3x+1\right)\sqrt{2x^2-1}=5x^2+\frac{3}{2}x-3\)ĐK \(2x^2-1\ge0\)

<=> \(10x^2-3x-6-2\left(3x+1\right)\sqrt{2x^2-1}=0\)

<=> \(7x^2-4x-8+\left(3x+1\right)\left(x+2-2\sqrt{2x^2-1}\right)=0\)

<=>\(7x^2-4x-8+\left(3x+1\right).\frac{\left(x+2\right)^2-4\left(2x^2-1\right)}{x+2+2\sqrt{2x^2-1}}=0\)

<=> \(7x^2-4x-8+\left(3x+1\right).\frac{-7x^2+4x+8}{x+2+2\sqrt{2x^2-1}}=0\)

<=>\(\orbr{\begin{cases}7x^2-4x-8=0\left(1\right)\\1-\frac{3x+1}{x+2+2\sqrt{2x^2-1}}=0\left(2\right)\end{cases}}\)

Giải (2)

\(2\sqrt{2x^2-1}=2x-1\)

<=> \(\hept{\begin{cases}x\ge\frac{1}{2}\\4x^2+4x-5=0\end{cases}}\)

=> \(x=\frac{-1+\sqrt{6}}{2}\)(thỏa mãn ĐKXĐ)

Giải (1)=> \(x=\frac{2+2\sqrt{15}}{7}\)

Vậy \(S=\left\{\frac{2+2\sqrt{15}}{7},\frac{-1+\sqrt{6}}{2}\right\}\)

\(P=a^2-2a+b^2-2b+c^2-2c+3\)

\(P=\left(a^2+\dfrac{9}{4}\right)+\left(b^2+4\right)+\left(c^2+\dfrac{25}{4}\right)-2a-2b-2c-\dfrac{19}{2}\)

\(P\ge3a+4b+5c-2a-2b-2c-\dfrac{19}{2}\)

\(P\ge a+2b+3c-\dfrac{19}{2}=13-\dfrac{19}{2}=\dfrac{7}{2}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};2;\dfrac{5}{2}\right)\)

24+t−94(∗)
Xét hàm (∗) được: MinF(t)=F(23)=−19
⇒MinP=MinF(t)=−19.dấu "=" xảy ra khi a=b=c=13