cho a,b,c>0. C/M a) \(\frac{_{-a+b+c}}{2a}+\frac{a-b+c}{2b}+\frac{a+b-c}{2c}\ge\frac{3}{2}\)\(\frac{3}{2}\)
b) \(\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\ge a+b+c\)
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NH
19 tháng 2 2020
Áp dụng bđt Cauchy-schwarz dạng engel ta có:
1. \(\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2a}\ge\frac{\left(a+b+c\right)^2}{\left(a+2b\right)+\left(b+2c\right)+\left(c+2a\right)}=\frac{a+b+c}{3}\)
Dấu "=" \(\Leftrightarrow\frac{a}{a+2b}=\frac{b}{b+2c}=\frac{c}{c+2a}\Leftrightarrow a=b=c\)
2. \(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{\left(2a+3b\right)+\left(2b+3c\right)+\left(2c+3a\right)}=\frac{a+b+c}{5}\)
Dấu "=" \(\Leftrightarrow a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 6 2019
a/ BĐT sai, cho \(a=b=c=2\) là thấy
b/ \(VT=\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{\left(a+b+c\right)^2}\)
\(VT\ge\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2}{3\left(a+b+c\right)^2}=\frac{1}{3}\left(a^2+b^2+c^2\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
c/ Tiếp tục sai nữa, vế phải là \(\frac{3}{2}\) chứ ko phải \(2\), và hy vọng rằng a;b;c dương
\(VT=\frac{a^2}{abc.b+a}+\frac{b^2}{abc.c+b}+\frac{c^2}{abc.a+c}\ge\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)+a+b+c}\)
\(VT\ge\frac{9}{3abc+3}\ge\frac{9}{\frac{3\left(a+b+c\right)^3}{27}+3}=\frac{9}{\frac{3.3^3}{27}+3}=\frac{9}{6}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
Ta có:
\(a^3+b^3+b^3\ge3ab^2\) ; \(b^3+c^3+c^3\ge3bc^2\) ; \(c^3+a^3+a^3\ge3ca^2\)
Cộng vế với vế \(\Rightarrow a^3+b^3+c^3\ge ab^2+bc^2+ca^2\)
\(\frac{a^5}{b^2}+\frac{b^5}{c^2}+\frac{c^5}{a^2}=\frac{a^6}{ab^2}+\frac{b^6}{bc^2}+\frac{c^6}{ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab^2+bc^2+ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=a^3+b^3+c^3\)
TN
1 tháng 4 2017
Bài 1:
\(BDT\Leftrightarrow\sqrt{\frac{3}{a+2b}}+\sqrt{\frac{3}{b+2c}}+\sqrt{\frac{3}{c+2a}}\le\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)
\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
Áp dụng BĐT Cauchy-Schwarz và BĐT AM-GM ta có:
\(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}\ge\frac{9}{\sqrt{a}+\sqrt{2}\cdot\sqrt{2b}}\ge\frac{9}{\sqrt{\left(1+2\right)\left(a+2b\right)}}=\frac{3\sqrt{3}}{\sqrt{a+2b}}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{c}}\ge\frac{3\sqrt{3}}{\sqrt{b+2c}};\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{a}}\ge\frac{3\sqrt{3}}{\sqrt{c+2a}}\)
Cộng theo vế 3 BĐT trên ta có:
\(3\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\ge3\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
Đẳng thức xảy ra khi \(a=b=c\)
Bài 2: làm mãi ko ra hình như đề sai, thử a=1/2;b=4;c=1/2
1 tháng 4 2017
Bài 2/
\(\frac{bc}{a^2b+a^2c}+\frac{ca}{b^2c+b^2a}+\frac{ab}{c^2a+c^2b}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2c^2b}+\frac{c^2a^2}{b^2c^2a+b^2a^2c}+\frac{a^2b^2}{c^2a^2b+c^2b^2a}\)
\(=\frac{b^2c^2}{ab+ac}+\frac{c^2a^2}{bc+ba}+\frac{a^2b^2}{ca+cb}\)
\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
\(\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3}{2}\)
Dấu = xảy ra khi \(a=b=c=1\)
DT
Cho a,b,c>0 CMR
\( \frac{a^3}{a+2b}+ \frac{b^3}{b+2c}+ \frac{c^3}{c+2a} \ge \frac{a^2+b^2+c^2}{3} \)
1
AH
Akai Haruma
Giáo viên
25 tháng 10 2017
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\text{VT}=\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2ac}\geq \frac{(a^2+b^2+c^2)^2}{a^2+2ab+b^2+2bc+c^2+2ac}\)
\(\Leftrightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{(a+b+c)^2}\) (1)
Theo hệ quả của BĐT AM-GM thì ta có:
\(a^2+b^2+c^2\geq ab+bc+ac\Leftrightarrow 3(a^2+b^2+c^2)\geq (a+b+c)^2\) (2)
Từ \((1),(2)\Rightarrow \text{VT}\geq \frac{a^2+b^2+c^2}{3}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c>0\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 4 2019
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
LH
Cho a,b,c>0.Chung minh rang \(\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\ge\frac{a+b+c}{3}\)
1
29 tháng 1 2020
Ta có:
\(\left(\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\right)\left[\left(b+2c\right)+\left(c+2a\right)+\left(a+2b\right)\right]\)
\(\ge\left[\sqrt{\frac{a^2}{b+2c}.\left(b+2\right)}+\sqrt{\frac{b^2}{c+2a}.\left(c+2a\right)}+\sqrt{\frac{c^2}{a+2b}.\left(a+2b\right)}\right]^2\)
\(=\left(a+b+c\right)^2\)
\(\Rightarrow\left(\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\right)\left[3\left(a+b+c\right)\right]\ge\left(a+b+c\right)^2\)
\(\Rightarrow\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\ge\frac{a+b+c}{3}\left(đpcm\right)\)
a, mk ko chép lại đề đâu nhé
=\(\frac{1}{2}\left(\frac{-a+b+c}{a}+\frac{a-b+c}{b}+\frac{a+b-c}{c}\right)\)
\(=\frac{1}{2}\left(-1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}-1+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}-1\right)\)
\(=\frac{1}{2}\left(-3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}\right)\)
Áp dụng BĐT Cô-si cho 2 số dương ta có
\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\)
\(\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{b}{c}.\frac{c}{b}}=2\)
\(\frac{a}{c}+\frac{c}{a}\ge2\sqrt{\frac{a}{c}.\frac{c}{a}}=2\)
=>\(\frac{1}{2}\left(-3+\frac{b}{a}+\frac{a}{b}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}\right)\)\(\ge\frac{1}{2}\left(-3+2+2+2\right)=\frac{3}{2}\)
=>dpcm