Áp dụng BĐT BSC:

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

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

\(\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{1}{2}\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Áp dụng bđt cosi schwart ta có:

`VT>=(a+b+c)^2/(a+b+c+sqrt{ab}+sqrt{bc}+sqrt{ca})`

Dễ thấy `sqrt{ab}+sqrt{bc}+sqrt{ca}<a+b+c`

`=>VT>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2=3`

Dấu "=" `<=>a=b=c=1.`

làm rõ \(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)

\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)

\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\) (đúng)

ok thỏa thuận rồi tui làm nửa sau thui nhé :D

Đặt \(a^2=x;b^2=y;c^2=z\) thì ta có:

\(VT=\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{y+z}}+\sqrt{\dfrac{z}{x+z}}\)

Lại có: \(\sqrt{\dfrac{x}{x+y}}=\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)

Tương tự cộng theo vế rồi áp dụng BĐT C-S ta có:

\(VT^2\le2\left(x+y+z\right)\left[\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(y+z\right)\left(y+x\right)}+\dfrac{z}{\left(z+x\right)\left(z+y\right)}\right]\)

\(\Leftrightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

Vì \(VP^2=\dfrac{9}{2}\) nên cần cm \(VT\le \frac{9}{2}\)

\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+yz+xz\right)\)

Can you continue

a)Bunhia:

\(\left(1+2\right)\left(b^2+2a^2\right)\ge\left(1.b+\sqrt{2}.\sqrt{2}a\right)^2=\left(b+2a\right)^2\)

b)\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng bđt câu a

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

\(\Leftrightarrow VT\ge\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}=3=VP\)

Tự tìm dấu "="

Nguyễn Việt LâmMashiro ShiinaBNguyễn Thanh HằngonkingCẩm MịcFa CTRẦN MINH HOÀNGhâu DehQuân Tạ MinhTrương Thị Hải Anh

\(1.\) Gỉa sử : \(\sqrt{25-16}< \sqrt{25}-\sqrt{16}\)

\(\Leftrightarrow3< 1\) ( Vô lý )

\(\Rightarrow\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)

\(2.\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)

\(\Leftrightarrow a-2\sqrt{ab}+b< a-b\)

\(\Leftrightarrow2b-2\sqrt{ab}< 0\)

\(\Leftrightarrow2\left(b-\sqrt{ab}\right)< 0\)

Ta có :\(a>b\Leftrightarrow ab>b^2\Leftrightarrow\sqrt{ab}>b\)

\(\RightarrowĐpcm.\)

\(2a.\) Áp dụng BĐT Cauchy , ta có :

\(a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\)

\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)

\(b.\) Áp dụng BĐT Cauchy cho các số dương , ta có :

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\left(x,y>0\right)\left(1\right)\)

\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(y,z>0\right)\left(2\right)\)

\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\left(x,z>0\right)\left(3\right)\)

Cộng từng vế của ( 1 ; 2 ; 3 ) , ta được :

\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)

\(3a.\sqrt{x-4}=a\left(a\in R\right)\left(x\ge4;a\ge0\right)\)

\(\Leftrightarrow x-4=a^2\)

\(\Leftrightarrow x=a^2+4\left(TM\right)\)

\(3b.\sqrt{x+4}=x+2\left(x\ge-2\right)\)

\(\Leftrightarrow x+4=x^2+4x+4\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)

KL....

Đặt vế trái là T, ta có:

\(\dfrac{a}{\sqrt{b+1}}=\dfrac{a\sqrt{2}}{\sqrt{2}.\sqrt{b+1}}\ge\dfrac{a\sqrt{2}}{\dfrac{b+1+2}{2}}=\dfrac{a.2\sqrt{2}}{b+3}\)

Tương tự: \(\dfrac{b}{\sqrt{c+1}}\ge\dfrac{b.2\sqrt{2}}{c+3}\)

\(\dfrac{c}{\sqrt{a+1}}\ge\dfrac{c.2\sqrt{2}}{a+3}\)

Cộng vế theo vế các BĐT vừa chứng minh, ta được

\(T\ge2\sqrt{2}\left(\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{a+3}\right)=2\sqrt{2}\left(\dfrac{a^2}{ab+3a}+\dfrac{b^2}{bc+3b}+\dfrac{c^2}{ac+3c}\right)\)

\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+3\left(a+b+c\right)}\)

\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{\dfrac{\left(a+b+c\right)^2}{3}+3\left(a+b+c\right)}\)

\(T\ge2\sqrt{2}.\dfrac{3^2}{\dfrac{3^2}{3}+9}=\dfrac{3\sqrt{2}}{2}\)(đpcm)

Đẳng thức xảy ra khi a=b=c=1

b) Đặt vế trái là N,ta có:

\(\sum\sqrt{\dfrac{a^3}{b+3}}=\sum\sqrt{\dfrac{a^4}{ab+3}}=\sum\dfrac{a^2}{\sqrt{ab+3}}=\sum\dfrac{2a^2}{\sqrt{4a\left(b+3\right)}}\ge\sum\dfrac{2a^2}{\dfrac{4a+b+3}{2}}=\sum\dfrac{4a^2}{4a+b+3}\)

\(\sum\dfrac{4a^2}{4a+b+3}\ge\dfrac{\left(2a+2b+2c\right)^2}{4a+b+3+4b+c+3+4c+a+3}=\dfrac{3}{2}\)(đpcm)

Đẳng thức xảy ra khi a=b=c=1

Bài 1:

Áp dụng BĐT AM-GM ta có:

$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$

Cộng theo vế và thu gọn:

$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$

Ta có đpcm.

Bài 2:

$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$

$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$

$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$

Cộng theo vế và rút gọn thu được:

$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$ 

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Bài 3:

Áp dụng BĐT Cauchy-Schwarz:

$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{b+c+c+a+a+b}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

Đặt vế trái BĐT cần chứng minh là P

Ta có:

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

Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2011}\)

Đồng thời: \(\left\{{}\begin{matrix}y^2+z^2-x^2=2a^2\\z^2+x^2-y^2=2b^2\\x^2+y^2-z^2=2c^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\right)\)

\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z+z+x+x+y\right)^2}{2x+2y+2z}-\left(x+y+z\right)\right)=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\sqrt{\dfrac{2011}{2}}\)