1: Cho a,b,c>0. CMR: \(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ac}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{c^2+a^2}{b^2+ac}\ge\dfrac{9}{2}\)
chứng minh BĐT
\(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\ge\dfrac{9}{2}\)
Áp dụng bất đẳng thức \(AM-GM\) cho 2 số dương ta có:
\(VT=\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\ge\dfrac{3abc}{2abc}+\dfrac{2ab}{c^2+ab}+\dfrac{2bc}{a^2+bc}+\dfrac{2ac}{b^2+ac}=\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\)
Áp dụng bất đẳng thức \(Cauchy-Schwarz\) \(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}=\dfrac{a^2b^2}{c^2ab+a^2b^2}+\dfrac{b^2c^2}{a^2bc+b^2c^2}+\dfrac{a^2c^2}{b^2ac+a^2c^2}\ge\dfrac{\left(ab+bc+ac\right)^2}{c^2ab+a^2b^2+a^2bc+b^2c^2+b^2ac+a^2c^2}\)
Đặt: \(\left\{{}\begin{matrix}ab=x\\bc=y\\ac=z\end{matrix}\right.\) ta được: \(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+xy+xz+xy}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\)
Nên: \(\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\ge\dfrac{3}{2}+2.\dfrac{3}{2}=\dfrac{9}{2}\)
Mà: \(VT\ge\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\Leftrightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)
Lời giải:
Áp dụng BĐT AM-GM ta có: \(\frac{a^3+b^3+c^3}{2abc}\geq \frac{3\sqrt[3]{a^3b^3c^3}}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\) (1)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{a^2+c^2}{b^2+ac}\geq \frac{(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2}{a^2+b^2+c^2+ab+bc+ac}\) (2)
Có:
\((\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2=2(a^2+b^2+c^2)+2\sqrt{(a^2+b^2)(b^2+c^2)}+2\sqrt{(b^2+c^2)(c^2+a^2)}+\sqrt{(a^2+b^2)(c^2+a^2)}\)
Áp dụng BĐT Bunhiacopxky:
\(\sqrt{(a^2+b^2)(b^2+c^2)}\geq \sqrt{(ac+b^2)^2}=ac+b^2\)
\(\sqrt{(b^2+c^2)(c^2+a^2)}\geq \sqrt{(ba+c^2)^2}=ba+c^2\)
\(\sqrt{(a^2+b^2)(c^2+a^2)}\geq \sqrt{(a^2+bc)^2}=a^2+bc\)
\(\Rightarrow (\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2\geq 2(a^2+b^2+c^2)+2(a^2+b^2+c^2+ab+bc+ac)\)
\(\geq a^2+b^2+c^2+ab+bc+ac+2(a^2+b^2+c^2+ab+bc+ac)\) (AM-GM)
Hay \((\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2\geq 3(a^2+b^2+c^2+ab+bc+ac)\) (3)
Từ \((2); (3)\Rightarrow \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{a^2+c^2}{b^2+ac}\geq 3\) (4)
Từ \((1); (4)\Rightarrow \frac{a^3+b^3+c^3}{2abc}+\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ac}\geq \frac{9}{2}\)
Ta có đpcm.
Dấu bằng xảy ra khi $a=b=c$
1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
CMR: Với các số thực dương a;b;c thì\(\dfrac{a^3+2abc+b^3}{c^2+ab}+\dfrac{a^3+2abc+c^3}{b^2+ac}+\dfrac{b^3+2abc+c^3}{a^2+bc}\ge2\left(a+b+c\right)\)
Cho a, b, c>0 và a+b+c\(\ge3\)
Cmr:
\(\dfrac{a^2}{a+\sqrt{bc}}+\dfrac{b^2}{b+\sqrt{ac}}+\dfrac{c^2}{c+\sqrt{ab}}\ge\dfrac{3}{2}\)
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.`
Cho a, b, c > . Chứng minh rằng:
a, \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
b, \(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
a.
Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
Cộng vế:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
b.
Ta có:
\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)
Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)
Cộng vế với vế:
\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{a+b+c}{2abc}\)
Dấu "=" xảy ra khi \(a=b=c\)
a ) Cho a,b,c >0 C/m:
\(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{a^2+b^2+c^2}{a+b+c}\)
b ) Cho a,b,c > 0 . C/m :
\(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge\dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.\)
c ) Cho a,b,c > 0 . C/m :
\(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge a+b+c.\)
giúp nha mn
a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)
\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ca+a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}\)
\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)
b)
Áp dụng BĐT Cauchy Shwarz, ta có:
\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2\)
Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:
\(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\dfrac{\left(a+b+c\right)^3}{9}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\dfrac{\left(a+b+c\right)}{3}\times\left(a^2+b^2+c^2\right)}\)
\(=\dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}\) (đpcm)
Dấu "=" xảy ra khi a = b = c.
Cho 3 số dương a,b,c
CMR : \(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(a+c\right)^2}\ge\dfrac{9}{4\left(ab+ac+bc\right)}\)
Đây là BĐT Iran 96 khá nổi tiếng. Bạn hoàn toàn có thể search trên google lời giải.
Cho a,b,c là số dương. CMR:
1. \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
2. \(a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\le a^3+b^3+c^3\)
3. \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
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 $a=b=c$
\(a,b,c>0.CMR:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
\(VT=\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2a\sqrt{bc}}+\dfrac{1}{2b\sqrt{ac}}+\dfrac{1}{2c\sqrt{ab}}\)
\(VT\le\dfrac{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}}{2abc}\)
Mặt khác ta luôn có:
\(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\)
\(\Rightarrow2\left(a+b+c\right)-2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)\ge0\)
\(\Rightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\le a+b+c\)
\(\Rightarrow VT\le\dfrac{a+b+c}{2abc}\)
Dấu "=" khi \(a=b=c\)