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

\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có : 

\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)

\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)

\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)

\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)

\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)

Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)

PS : Thánh cx đc phết ha; chế đc bài này tui mới khâm phục :)))

nó ko chém đâu anh nó chép trong toán tuổi thơ đấy,thk này khốn nạn lắm

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)

\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

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

\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)

Từ ( 1 ) và ( 2 ) có đpcm

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)thì \(x,y,z>0\)và ta cần chứng minh \(\frac{x}{\sqrt{3zx+yz}}+\frac{y}{\sqrt{3xy+zx}}+\frac{z}{\sqrt{3yz+xy}}\ge\frac{3}{2}\)\(\Leftrightarrow\frac{x^2}{x\sqrt{3zx+yz}}+\frac{y^2}{y\sqrt{3xy+zx}}+\frac{z^2}{z\sqrt{3yz+xy}}\ge\frac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng phân thức, ta có: \(\frac{x^2}{x\sqrt{3zx+yz}}+\frac{y^2}{y\sqrt{3xy+zx}}+\frac{z^2}{z\sqrt{3yz+xy}}\ge\)\(\frac{\left(x+y+z\right)^2}{x\sqrt{3zx+yz}+y\sqrt{3xy+zx}+z\sqrt{3yz+xy}}\)

Áp dụng BĐT Cauchy-Schwarz, ta có: \(x\sqrt{3zx+yz}+y\sqrt{3xy+zx}+z\sqrt{3yz+xy}\)\(=\sqrt{x}.\sqrt{3zx^2+xyz}+\sqrt{y}.\sqrt{3xy^2+xyz}+\sqrt{y}.\sqrt{3yz^2+xyz}\)\(\le\sqrt{\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]}\)

Ta cần chứng minh \(\sqrt{\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]}\le\frac{2}{3}\left(x+y+z\right)^2\)

\(\Leftrightarrow\left(x+y+z\right)^4\ge\frac{9}{4}\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]\)

\(\Leftrightarrow\left(x+y+z\right)^3\ge\frac{27}{4}\left(xy^2+yz^2+zx^2+xyz\right)\)(*)

Không mất tính tổng quát, giả sử \(y=mid\left\{x,y,z\right\}\)thì khi đó \(\left(y-x\right)\left(y-z\right)\le0\Leftrightarrow y^2+zx\le xy+yz\)

\(\Leftrightarrow xy^2+zx^2\le x^2y+xyz\Leftrightarrow xy^2+yz^2+zx^2+xyz\le\)\(x^2y+yz^2+2xyz=y\left(z+x\right)^2=4y.\frac{z+x}{2}.\frac{z+x}{2}\)

\(\le\frac{4}{27}\left(y+\frac{z+x}{2}+\frac{z+x}{2}\right)^3=\frac{4\left(x+y+z\right)^3}{27}\)

Như vậy (*) đúng

Đẳng thức xảy ra khi a = b = c

1.

\(\left(a+c\right)\left(b+c\right)=4c^2\Leftrightarrow\left(\frac{a}{c}+1\right)\left(\frac{b}{c}+1\right)=4\)

Đặt \(\left(\frac{a}{c};\frac{b}{c}\right)=\left(x;y\right)\Rightarrow\left(x+1\right)\left(y+1\right)=4\)

\(\Leftrightarrow xy+x+y=3\Rightarrow\left\{{}\begin{matrix}xy=3-\left(x+y\right)\\4\le\frac{1}{4}\left(x+y+2\right)^2\Rightarrow x+y\ge2\end{matrix}\right.\)

Ta có:

\(P=\frac{\frac{a}{c}}{\frac{b}{c}+3}+\frac{\frac{b}{c}}{\frac{a}{c}+3}+\frac{1}{\frac{c}{a}+\frac{c}{b}}=\frac{x}{y+3}+\frac{y}{x+3}+\frac{1}{\frac{1}{x}+\frac{1}{y}}=\frac{x}{y+3}+\frac{y}{x+3}+\frac{xy}{x+y}\)

\(P=\frac{x^2+y^2+3\left(x+y\right)}{\left(x+1\right)\left(y+1\right)+2\left(x+y\right)+8}+\frac{xy}{x+y}=\frac{\left(x+y\right)^2-2xy+3\left(x+y\right)}{2\left(x+y\right)+12}+\frac{xy}{x+y}\)

Đặt \(x+y=t\Rightarrow2\le t< 3\Rightarrow P=\frac{t^2-2\left(3-t\right)+3t}{2t+12}+\frac{3-t}{t}=\frac{t^2+5t-6}{2\left(t+6\right)}+\frac{3}{t}-1\)

\(P=\frac{\left(t-1\right)\left(t+6\right)}{2\left(t+6\right)}+\frac{3}{t}-1=\frac{t}{2}+\frac{3}{t}-\frac{1}{2}\)

\(P\ge2\sqrt{\frac{3t}{2t}}-\frac{1}{2}=\sqrt{6}-\frac{1}{2}\)

\(P_{min}=\sqrt{6}-\frac{1}{2}\) khi \(t=\sqrt{6}\)

\(P=\frac{t}{2}+\frac{3}{t}-\frac{1}{2}=\frac{t^2-5t+6}{2t}+2=\frac{\left(t-2\right)\left(t-3\right)}{2t}+2\le2\)

\(P_{max}=2\) khi \(t=2\)

2.

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

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

Ta có:

\(x+y=-z\Leftrightarrow\left(x+y\right)^5=-z^5\)

\(\Leftrightarrow x^5+y^5+z^5=-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)

\(\Leftrightarrow x^5+y^5+z^5=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)

\(\Leftrightarrow P=-5xy\left[\left(x+y\right)^3-xy\left(x+y\right)\right]\)

\(\Leftrightarrow P=-5xy\left[-z^3+xyz\right]=5xyz\left(z^2-xy\right)\)

\(\Leftrightarrow P=\frac{5}{2}xyz\left(z^2+\left(x+y\right)^2-2xy\right)=\frac{5}{2}xyz\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow P=\frac{5}{2}xyz=\frac{5}{2}x\left(x^2-\frac{1}{2}\right)\)

\(\Rightarrow P^2=\frac{25}{4}x^2\left(\frac{1}{2}-x^2\right)^2=\frac{25}{8}.2x^2\left(\frac{1}{2}-x^2\right)\left(\frac{1}{2}-x^2\right)\)

\(\Rightarrow P^2\le\frac{25}{8}\left(\frac{2x^2+\frac{1}{2}-x^2+\frac{1}{2}-x^2}{3}\right)^3=\frac{25}{216}\)

\(\Rightarrow P\le\frac{5\sqrt{6}}{36}\)

\(P_{max}=\frac{5\sqrt{6}}{36}\) khi \(x=-\frac{1}{\sqrt{6}}\)

3.

Xét \(Q=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)

\(Q^2=\frac{a^4}{b^2}+\frac{2a^2b}{c}+c^2+\frac{b^4}{c^2}+\frac{2b^2c}{a}+a^2+\frac{c^4}{a^2}+\frac{2c^2a}{b}+b^2-\left(a^2+b^2+c^2\right)\)

\(\Rightarrow Q^2\ge4\sqrt[4]{\frac{a^4.a^2b.a^2b.c^2}{b^2c^2}}+4\sqrt[4]{\frac{b^4.b^2c.c^2c.a^2}{c^2a^2}}+4\sqrt[4]{\frac{c^4.c^2a.c^2a.b^2}{a^2b^2}}-\left(a^2+b^2+c^2\right)\)

\(\Rightarrow Q^2\ge3\left(a^2+b^2+c^2\right)\Rightarrow Q\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)

Đặt \(x=a^2+b^2+c^2\ge\frac{1}{3}\)

\(\Rightarrow P\ge2020\sqrt{3x}+\frac{1}{3x}=\sqrt{3x}+\sqrt{3x}+\frac{1}{3x}+2018\sqrt{3x}\)

\(\Rightarrow P\ge3\sqrt[3]{\frac{3x}{3x}}+2018.\sqrt{3.\frac{1}{3}}=2021\)

\(P_{min}=2021\) khi \(a=b=c=\frac{1}{3}\)

\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)

\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)

\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)

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

\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

Vậy VT = VP, đẳng thức được chứng minh

ko khó nhưng mà bn đăng từng câu 1 hộ mk mk giải giúp cho

gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

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

=> Thay vào thì     \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)

\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)

Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào

=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)

=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)

=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\) 

Đặt: \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)

=>     \(P=\frac{xy}{z^2+3xy}+\frac{yz}{x^2+3yz}+\frac{zx}{y^2+3zx}\)

=>     \(3P=\frac{3xy}{z^2+3xy}+\frac{3yz}{x^2+3yz}+\frac{3zx}{y^2+3zx}=1-\frac{z^2}{z^2+3xy}+1-\frac{x^2}{x^2+3yz}+1-\frac{y^2}{y^2+3zx}\)

Ta sẽ CM: \(3P\le\frac{9}{4}\)<=> Cần CM: \(\frac{x^2}{x^2+3yz}+\frac{y^2}{y^2+3zx}+\frac{z^2}{z^2+3xy}\ge\frac{3}{4}\)

Có:    \(VT\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)

Ta sẽ CM: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{3}{4}\)

<=> \(4\left(x+y+z\right)^2\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(4\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(x^2+y^2+z^2\ge xy+yz+zx\)

Mà đây lại là 1 BĐT luôn đúng => \(3P\le\frac{9}{4}\)=> \(P\le\frac{3}{4}\)

Vậy P max \(=\frac{3}{4}\)<=> \(a=b=c\)

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

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

\(=\left(\frac{-3a+2b+c+2a+b-3c+a-3b+2c}{6}\right)^3-\frac{\left(-a+3b-2c\right)\left(3a-2b-c\right)\left(-2a-b+3c\right)}{72}\)

(Hằng đẳng thức)

\(=0-\frac{\left(-a+3b-2c\right)\left(3a-2b-c\right)\left(-2a-b+3c\right)}{72}\)

\(\Rightarrow\frac{\left(a-3b+2c\right)\left(-3a+2b+c\right)\left(2a+b-3c\right)}{72}=\frac{1}{8}\)

\(\Leftrightarrow\left(a-3b+2c\right)\left(2a+b-3c\right)\left(-3a+2b+c\right)=9\)(đpcm).

\(\frac{1}{a}\ge1-\frac{2}{2b+1}+1-\frac{3}{3c+2}=\frac{2b-1}{2b+1}+\frac{3c-1}{3c+2}\ge2\sqrt{\frac{\left(2b-1\right)\left(3c-1\right)}{\left(2b+1\right)\left(3c+2\right)}}\)

Tương tự: \(\frac{2}{2b+1}\ge\frac{a-1}{a}+\frac{3c-1}{3c+2}\ge2\sqrt{\frac{\left(a-1\right)\left(3c-1\right)}{a\left(3c+2\right)}}\)

\(\frac{3}{3c+2}\ge\frac{a-1}{a}+\frac{2b-1}{2b+1}\ge2\sqrt{\frac{\left(a-1\right)\left(2b-1\right)}{a\left(2b+1\right)}}\)

Nhân vế với vế:

\(\frac{6}{a\left(2b+1\right)\left(3c+2\right)}\ge\frac{8\left(a-1\right)\left(2b-1\right)\left(3c-1\right)}{a\left(2b+1\right)\left(3c+2\right)}\)

\(\Rightarrow\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\le\frac{3}{4}\)