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 $a=b=c$

tìm trc khi hỏi Câu hỏi của Hoàng Thiên - Toán lớp 9 - Học toán với OnlineMath

bạn ầy làm đúng rồi

k tui nha

thank

bạn áp dụng Bunhiacopxki là ra mà : VT^2 <= ( c +b -c ). ( a - c + c ) = ab

sửa đề\(\frac{1}{x^2+1}+\frac{1}{y^2+1}\ge\frac{2}{1+xy}\)

\(\Leftrightarrow\frac{1}{x^2+1}+\frac{1}{y^2+1}-\frac{2}{1+xy}\ge0\)

\(\Leftrightarrow\left(\frac{1}{1+x^2}-\frac{1}{1+xy}\right)+\left(\frac{1}{1+y^2}-\frac{1}{1+xy}\right)\ge0\)

\(\Leftrightarrow\frac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\frac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow\frac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)( luôn đúng với \(x,y\ge1\))

Đpcm

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

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

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( 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

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

Bđt Bu-nhia-cop-xki \(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\), đẳng thức xảy ra khi \(ay=bx\)

a.

\(\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\le\left(2+3\right)\left(2x^2+3y^2\right)=5^2\)

\(\Rightarrow-5\le2x+3y\le5\)

b.

\(\sqrt{a+c}.\sqrt{b+c}+\sqrt{a-c}.\sqrt{b-c}\le\sqrt{a+c+a-c}.\sqrt{b+c+b-c}\)

\(=\sqrt{2a}.\sqrt{2b}=2\sqrt{ab}\)

Dấu bằng xảy ra khi \(\frac{\sqrt{a+c}}{\sqrt{a-c}}=\frac{\sqrt{b+c}}{\sqrt{b-c}}\), hay \(a=b\)

Thử lại với a = b thì \(VT=2a=2\sqrt{ab}=VP>\sqrt{ab}\) nên đề đã ra sai vế phải của bđt.

c.

bđt \(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)

d.

bđt \(\Leftrightarrow\left(a+c\right)^2+\left(b+d\right)^2\le a^2+b^2+c^2+d^2+2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\)

\(\Leftrightarrow ac+bd\le\sqrt{a^2+b^2}.\sqrt{c^2+d^2}\)

bđt trên luôn đúng vì theo bđt Bu-nhia-cop-xki, ta có:

\(\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\sqrt{\left(ac+bd\right)^2}=\left|ac+bd\right|\ge ac+bd\)