\(3=a+b+ab\le a+b+\frac{\left(a+b\right)^2}{4}\Rightarrow\left(a+b\right)^2+4\left(a+b\right)-12\ge0\)

\(\Leftrightarrow\left(a+b-2\right)\left(a+b+6\right)\ge0\Rightarrow a+b\ge2\)

Đặt vế trái của BĐT là P

\(P=\frac{4a\left(a+1\right)+4b\left(b+1\right)}{\left(a+1\right)\left(b+1\right)}+2ab-\sqrt{7-3\left(3-a-b\right)}\)

\(P=\frac{4\left(a^2+b^2+a+b\right)}{ab+a+b+1}+2ab-\sqrt{3\left(a+b\right)-2}\)

\(P=a^2+b^2+a+b+2ab-\sqrt{3\left(a+b\right)-2}\)

\(P=\left(a+b\right)^2+a+b-\sqrt{3\left(a+b\right)-2}\)

Đặt \(\sqrt{3\left(a+b\right)-2}=x\Rightarrow\left\{{}\begin{matrix}x\ge2\\a+b=\frac{x^2+2}{3}\end{matrix}\right.\)

\(\Rightarrow P=\left(\frac{x^2+2}{3}\right)^2+\frac{x^2+2}{3}-x=\frac{x^4+7x^2-9x+10}{9}\)

\(P=\frac{x^4+7x^2-9x-26+36}{9}=\frac{\left(x-2\right)\left(x^3+2x^2+11x+13\right)}{9}+4\ge4\) ; \(\forall x\ge2\) (đpcm)

Dấu "=" xảy ra khi \(x=2\) hay \(a=b=1\)

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 \(=\frac{3}{4}\)<=> \(a=b=c\)

Bài 1:

Đặt \(a^2=x;b^2=y;c^2=z\)

Ta có:\(\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\le\frac{3}{\sqrt{2}}\)

Áp dụng BĐT cô si ta có:

\(\sqrt{\frac{x}{x+y}}=\frac{1}{\sqrt{2}}\sqrt{\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}\frac{3\left(x+z\right)}{2\left(x+y+z\right)}}\)

\(\le\frac{1}{2\sqrt{2}}\left[\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}+\frac{3\left(x+z\right)}{2\left(x+y+z\right)}\right]\)

Tương tự với \(\sqrt{\frac{y}{y+z}}\)và \(\sqrt{\frac{z}{z+x}}\)

Cộng lại ta được:

\(\frac{\sqrt{2}}{3}\left[\frac{x\left(x+y+z\right)}{\left(x+y\right)\left(x+z\right)}+\frac{y\left(x+y+z\right)}{\left(y+z\right)\left(y+x\right)}+\frac{z\left(x+y+z\right)}{\left(z+x\right)\left(z+y\right)}\right]+\frac{3}{2\sqrt{2}}\le\frac{3}{2\sqrt{2}}\)

Sau đó bình phương hai vế rồi

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)đẳng thức đúng

Vậy...

Bài 2:

Trước hết ta chứng minh bất đẳng thức sau:

\(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\le\frac{1}{3}\)

Nhân cả hai vế bđt với 4(a+b+c)4(a+b+c) rồi thu gọn ta được bđt sau: 

\(\frac{4a\left(a+b+c\right)}{4a+4b+c}+\frac{4b\left(a+b+c\right)}{4b+4c+a}+\frac{4c\left(a+b+c\right)}{4c+4a+b}\)\(\le\frac{4}{3}\left(a+b+c\right)\)

\(\left[\frac{4a\left(a+b+c\right)}{4a+4b+}-a\right]+\left[\frac{4b\left(a+b+c\right)}{4b+4c+a}-b\right]+\left[\frac{4c\left(a+b+c\right)}{4c+4a+b}-c\right]\le\frac{a+b+c}{3}\)

\(\frac{ca}{4a+4b+c}+\frac{ab}{4b+4c+a}+\frac{bc}{4c+4a+b}\le\frac{a+b+c}{9}\)

Áp dụng bđt cauchy-Schwarz ta có \(\frac{ca}{4a+4b+c}=\frac{ca}{\left(2b+c\right)+2\left(2a+b\right)}\)\(\le\frac{ca}{9}\left(\frac{1}{2b+c}+\frac{2}{2a+b}\right)\)

Từ đó ta có:

\(\text{∑}\frac{ca}{4a+4b+c}\le\frac{1}{9}\text{∑}\left(\frac{ca}{2b+c}+\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ab}{2b+c}\right)=\frac{a+b+c}{9}\)

Đặt VT=A rồi áp dụng bđt cauchy-Schwarz cho VT ta có 

\(T^2\le3\left(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\right)\)\(\le3\cdot\frac{1}{3}=1\Leftrightarrow T\le1\)

Dấu = xảy ra khi a=b=c 

c bạn tự làm nhé mình mệt rồi :D

- Ôi má ơi, má patient dử dậy :)

Trả lời hộ mình đi

ĐÂY MÀ LÀ toán 5 ạ??

Gọi A là vế trái của BĐT cần chứng minh. Không mất tính tổng quát, ta giả sử a + b + c = 3. Áp dụng BĐT AM - GM ta có:

\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{8bc\left(4a+4b+c\right)}}+\frac{ab\left(4a+4b+c\right)}{27}\)\(\ge\frac{1}{2}\left(a+b\right)\)

Suy ra 

             \(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}\)\(+\frac{ab\left(4a+4b+c\right)}{54}\ge\frac{1}{4}\left(a+b\right)\)

Tương tự

            \(\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\frac{bc\left(4b+4c+a\right)}{54}\ge\frac{1}{4}\left(b+c\right)\)

và       \(\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}+\frac{ca\left(4c+4a+b\right)}{54}\ge\frac{1}{4}\left(c+a\right)\)

Cộng ba BĐT trên ta có: 

           \(\frac{1}{2\sqrt{2}}A\ge B\)

Với \(A=\frac{1}{54}[ab\left(4a+4b+c\right)+bc\left(4b+4c+a\right)\)

\(+ca\left(4c+4a+b\right)]\)

\(=\frac{1}{54}\left[4ab\left(a+b\right)+4bc\left(b+c\right)+4ca\left(c+a\right)+3abc\right]\)

\(=\frac{1}{54}\left[4\left(a+b+c\right)\left(ab+bc+ca\right)-9abc\right]\)

\(\le\frac{1}{54}\left(a+b+c\right)^3=\frac{1}{2}\)

và \(B=\frac{1}{4}.2\left(a+b+c\right)=\frac{3}{2}\)

Suy ra \(\frac{1}{2\sqrt{2}}A\ge\frac{3}{2}-\frac{1}{2}=1\Rightarrow A\ge2\sqrt{2}\)

Vậy 

              \(\sqrt{\frac{\left(a+b\right)^3}{ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{bc\left(4a+4b+c\right)}}+\sqrt{\frac{\left(c+a\right)^3}{ca\left(4c+4a+b\right)}}\ge2\sqrt{2}\)(đpcm)

toán lớp 5 phiên bản hack não

Cái này không khó :v

Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:

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

Face khác ;v, theo AM-GM, ta có

\(\dfrac{a+b+c}{2}\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{6}{2}=3\)

Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c=2

Thôi giải lại câu 1:v (ý tưởng dồn biến là quá trâu bò! Bên AoPS em mới phát hiện ra có một cách bằng Cauchy-Schwarz quá hay!)

\(BĐT\Leftrightarrow\Sigma_{cyc}\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{9}{2}\)(*)

BĐT này đúng theo Cauchy-Schwarz: \(VT_{\text{(*)}}\le\Sigma_{cyc}\left(\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{9}{2}\)

Ta có đpcm.

Equality holds when a = b = c = 1 (Đẳng thức xảy ra khi a = b =c = 1)

1/Đặt \(VT=f\left(a;b;c\right)\) và \(0< t=\frac{a+b}{2}\)

Ta có: \(f\left(a;b;c\right)-f\left(t;t;c\right)=\frac{1}{4a^2+b^2+c^2}+\frac{1}{4b^2+a^2+c^2}-\frac{2}{5t^2+c^2}+\frac{1}{a^2+b^2+4c^2}-\frac{1}{2t^2+4c^2}\)

\(=\frac{5t^2-4a^2-b^2}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}+\frac{5t^2-4b^2-a^2}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{2t^2-a^2-b^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)

\(=-\frac{1}{4}\left(a-b\right)\left[\frac{\left(11a+b\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{\left(a+11b\right)}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}\right]+\frac{2t^2-a^2-b^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)

Xét cái ngoặc to: \(\frac{\left(11a+b\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{\left(a+11b\right)}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)

\(=\frac{\left(11a+b\right)\left(4b^2+a^2+c^2\right)-\left(a+11b\right)\left(4a^2+b^2+c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)

\(=\frac{\left(a-b\right)\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)

Từ đó: f(a;b;c) -f(t;t;c)

\(=-\frac{\frac{1}{4}\left(a-b\right)^2\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{-\frac{1}{2}\left(a-b\right)^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)

\(=-\frac{1}{4}\left(a-b\right)^2\left[\frac{\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\right]\le0\)

Do đó \(f\left(a;b;c\right)\le f\left(t;t;c\right)=f\left(t;t;3-2t\right)\)

\(=\frac{-9\left(t-1\right)^4}{2\left(3t^2-8t+6\right)\left(3t^2-4t+3\right)}+\frac{1}{2}\le\frac{1}{2}\)

Ta có đpcm.