Cho a,b dương. CM \(\frac{a^2+b^2}{a+b}\ge\frac{a+b}{2}\)
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Q
cho a,b,c, dương tm abc =8 . CM\(\frac{a+b+c}{2}\ge\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}\)
0
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2020
\(VT\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)}\ge\frac{3\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\) (1)
Mặt khác:
\(\left(a+b+c\right)^2=\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)\ge3\sqrt[3]{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2}\)
\(\Leftrightarrow\left(a+b+c\right)^6\ge27\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^6}{27\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\Leftrightarrow\frac{\left(a+b+c\right)^2.3^4}{27\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\)
\(\Leftrightarrow\frac{3\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\) (2)
(1);(2) \(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge a^2+b^2+c^2\)
10 tháng 10 2019
Áp dụng BĐT Svác ta có:
\(\frac{a^2}{2b+c}+\frac{b^2}{2c+a}+\frac{c^2}{2a+b}\ge\frac{\left(a+b+c\right)^2}{3\left(a+b+c\right)}=\frac{a+b+c}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 5 2019
\(VT=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\ge\frac{1}{2}\left(\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\right)=\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\)
Dấu "=" xảy ra khi \(a=b=c\)
AH
Akai Haruma
Giáo viên
29 tháng 9 2019
Lời giải:
Ta có:
\(\text{VT}=2a-\frac{2ab^2}{a+b^2}+2b-\frac{2bc^2}{b+c^2}+2c-\frac{2ca^2}{c+a^2}\)
\(=2(a+b+c)-2\left(\frac{ab^2}{a+b^2}+\frac{bc^2}{b+c^2}+\frac{ca^2}{c+a^2}\right)(*)\)
Áp dụng BĐT AM-GM:
\(\frac{ab^2}{a+b^2}+\frac{bc^2}{b+c^2}+\frac{ca^2}{c+a^2}\leq \frac{ab^2}{2\sqrt{ab^2}}+\frac{bc^2}{2\sqrt{bc^2}}+\frac{ca^2}{2\sqrt{ca^2}}=\frac{\sqrt{ab^2}}{2}+\frac{\sqrt{bc^2}}{2}+\frac{\sqrt{ca^2}}{2}\)
\(\leq \frac{ab+b}{4}+\frac{bc+c}{4}+\frac{ca+a}{4}=\frac{ab+bc+ac+a+b+c}{4}(1)\)
Mà:
\((a+b+c)^2\geq 3(ab+bc+ac)=(a^2+b^2+c^2)(ab+bc+ac)\geq (ab+bc+ac)^2\)
\(\Rightarrow a+b+c\geq ab+bc+ac(2)\)
Từ \((1);(2)\Rightarrow \frac{ab^2}{a+b^2}+\frac{bc^2}{b+c^2}+\frac{ca^2}{c+a^2}\leq \frac{a+b+c}{2}(**)\)
Từ $(*); (**)\Rightarrow \text{VT}\geq a+b+c$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
N
29 tháng 9 2019
CÁCH KHÁC:
Áp dụng BĐT Svarxo:
\(VT=2\left(\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\right)\)\(\ge\frac{2\left(a+b+c\right)^2}{a+b+c+a^2+b^2+c^2}\)\(=\frac{2\left(a+b+c\right)^2}{a+b+c+3}\ge a+b+c\)
\(\Leftrightarrow2\left(a+b+c\right)^2-\left(a+b+c\right)\left(a+b+c+3\right)\ge0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)
Đặt t=a+b+c(t>0)
\(\Rightarrow t\left(t-3\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}t\le0\\t\ge3\end{matrix}\right.\)
Giả sử t<3 hay a+b+c<3
=> Mỗi số a,b,c<3
Bí rồi
15 tháng 8 2016
Theo bất đẳng thức Cauchy-Schwarzt ta có \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}.\)
Mặt khác, \(a^2+b^2+c^2\ge ab+bc+ca\), do đó ta suy ra \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2.\)
15 tháng 8 2016
P=\(\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}=a^2+b^2+c^2\)
27 tháng 2 2020
\(\Sigma\left(\frac{a^3}{a^2+b^2}\right)=\Sigma\left(\frac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}\right)=\Sigma\left(a-\frac{ab^2}{a^2+b^2}\right)\ge\Sigma\left(a-\frac{ab^2}{2ab}\right)=\Sigma\left(a-\frac{b}{2}\right)\)
\(=a+b+c-\left(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)=\frac{a+b+c}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 6 2019
\(a+2b=1.a+\sqrt{2}.\sqrt{2}b\le\sqrt{\left(1+2\right)\left(a^2+2b^2\right)}\le\sqrt{3.3c^2}=3c\)
\(\Rightarrow a+2b\le3c\)
\(\Rightarrow\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\) (đpcm)
Dấu "=" khi \(a=b=c\)
26 tháng 4 2019
\(3-B=\left(a-\frac{a}{1+b^2}\right)+\left(b-\frac{b}{1+c^2}\right)+\left(c-\frac{c}{1+a^2}\right)=\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2}+\frac{a^2}{1+a^2}\le\frac{b^2}{2b}+\frac{c^2}{2c}+\frac{a^2}{2a}=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
=> \(B\ge\frac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1
Y
26 tháng 4 2019
\(B=\frac{a\left(b^2+1\right)-ab^2}{b^2+1}+\frac{b\left(c^2+1\right)-bc^2}{c^2+1}+\frac{c\left(a^2+1\right)-ca^2}{c^2+1}\)
\(\Leftrightarrow B=a-\frac{ab^2}{b^2+1}+b-\frac{bc^2}{c^2+1}+c-\frac{ca^2}{a^2+1}\)
\(\Leftrightarrow B=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\)
+ \(b^2+1\ge2b\forall b\)
\(\Rightarrow\frac{ab^2}{b^2+1}\le\frac{ab^2}{2b}=\frac{ab}{2}\). Dấu "=" xảy ra \(\Leftrightarrow b=1\)
+ Tương tự ta cm đc :
\(\frac{bc^2}{c^2+1}\le\frac{bc}{2}\) . Dấu "=" xảy ra \(\Leftrightarrow c=1\)
\(\frac{ca^2}{a^2+1}\le\frac{ca}{2}\). Dấu '=" xảy ra \(\Leftrightarrow a=1\)
Do đó : \(\frac{ab^2}{a^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\le\frac{ab+bc+ca}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
+ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(a+bc+ca\right)\)
\(\Leftrightarrow ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Do đó : \(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\le\frac{ab+bc+ca}{2}\le\frac{3}{2}\)
\(\Leftrightarrow-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\ge-\frac{3}{2}\)
\(\Leftrightarrow B=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\)
\(\ge3-\frac{3}{2}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
\(\frac{a^2+b^2}{a+b}\ge\frac{a+b}{2}\)
\(\Leftrightarrow\frac{a^2+b^2}{a+b}-\frac{a+b}{2}\ge0\)
\(\Leftrightarrow\frac{2a^2+2b^2-\left(a+b\right)^2}{2.\left(a+b\right)}\ge0\)
\(\Leftrightarrow\frac{2a^2+2b^2-a^2-2ab-b^2}{2.\left(a+b\right)}\ge0\)
\(\Leftrightarrow\frac{a^2+b^2-2ab}{2.\left(a+b\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{2.\left(a+b\right)}\ge0\)(luôn đúng)
\(\Rightarrow\frac{a^2+b^2}{a+b}\ge\frac{a+b}{2}\)
đpcm
Ta có a,b>0 ; \(\frac{a^2+b^2}{a+b}\ge\frac{a+b}{2}\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
Vậy \(\frac{a^2+b^2}{a+b}\ge\frac{a+b}{2}\)