Bài 1:

Áp dụng BĐT Bunhiacopxky:

$(\sqrt{c(a-c)}+\sqrt{c(b-c)})^2\leq [c+(b-c)][(a-c)+c]=ab$

$\Rightarrow \sqrt{c(a-c)}+\sqrt{c(b-c)}\leq \sqrt{ab}$

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=2c$

Bài 2:

Áp dụng BĐT Bunhiacopkxy:

\((a\sqrt{b-1}+b\sqrt{a-1})^2=(\sqrt{a}.\sqrt{ab-a}+\sqrt{b}.\sqrt{ab-b})^2\)

\(\leq (a+b)(ab-a+ab-b)=(a+b)(2ab-a-b)\)

Áp dụng BĐT AM-GM:

$(a+b)(2ab-a-b)\leq \left(\frac{a+b+2ab-a-b}{2}\right)^2=(ab)^2$

Do đó:

$(a\sqrt{b-1}+b\sqrt{a-1})^2\leq (ab)^2$

$\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\leq ab$

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=2$

Lời giải:

Ta có:

\(A=(a+1)^2+\left(\frac{a^2+2a+2}{a+1}\right)^2=(a+1)^2+\left[\frac{(a+1)^2+1}{a+1}\right]^2\)

Đặt $a+1=t(t\neq 0)$ thì:

$A=t^2+(\frac{t^2+1}{t})^2=t^2+(t+\frac{1}{t})^2$

$=2t^2+\frac{1}{t^2}+2\geq 2\sqrt{2t^2.\frac{1}{t^2}}+2=2\sqrt{2}+2$ theo BĐT AM-GM

Vậy $A_{\min}=2\sqrt{2}+2$

Giá trị này đạt được khi $t=\frac{\pm 1}{\sqrt[4]{2}}$

$\Leftrightarrow a=\frac{\pm 1}{\sqrt[4]{2}}-1$

Bài 1:  (không dùng Cô-si) Bình phương hai vế, ta được:

\(c\left(a-c\right)+c\left(b-c\right)+2c\sqrt{\left(a-c\right)\left(b-c\right)}\le ab\)

\(ac-2c^2+bc+2c\sqrt{\left(a-c\right)\left(b-c\right)}\le ab\)

\(0\le\left(ab-ac-bc+c^2\right)+2c\sqrt{\left(a-c\right)\left(b-c\right)}+c^2\)

\(0\le\left(a-c\right)\left(b-c\right)+2c\sqrt{\left(a-c\right)\left(b-c\right)}+c^2\)

\(0\le\left(\sqrt{\left(a-c\right)\left(b-c\right)}-c\right)^2\)(đúng)

Vậy BĐT đúng.  Xảy ra khi  \(a=b=2c\)

\(\sqrt{\left(1+a\right)\left(1+b\right)}\ge1+\sqrt{ab}\)

\(\Leftrightarrow\left(1+a\right)\left(1+b\right)\ge\left(\sqrt{ab}+1\right)^2\)

\(\Leftrightarrow ab+a+b+1\ge ab+2\sqrt{ab}+1\)

\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow a=b>0\)

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OMG !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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 $a=b=c$

Nguyễn Thị Ngọc Thơ, Nguyễn Việt Lâm, @No choice teen, @Trần Thanh Phương, @Akai Haruma

giúp e vs ạ! Cần gấp!

thanks nhiều!