Cho a,b,c tm \(ab+bc+ca\le3abc\)
Cm \(\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\ge\frac{3}{2}\)
Cho 3 số dương a,b,c tm: a+b+c+ab+ca+bc=6abc
CMR: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{3}\)
Sửa \(\dfrac{1}{3}\rightarrow3\)
Từ \(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=6\)
Ta có: \(\dfrac{1}{a^2}+1\ge\dfrac{2}{a};\dfrac{1}{b^2}+1\ge\dfrac{2}{b};\dfrac{1}{c^2}+1\ge\dfrac{2}{c}\)
Và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{2}{ab};\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2}{bc};\dfrac{1}{c^2}+\dfrac{1}{a^2}\ge\dfrac{2}{ac}\)
Cộng theo vế các BĐT trên ta có:
\(3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge12\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\ge4\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\)
\("="\Leftrightarrow a=b=c=1\)
cho a,b>0 cm\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\) nếu \(ab\ge1\)
b) cho a,b,c\(\ge\)1. CMR \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)
\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)
b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)
\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)
Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)
Cộng vế với vế ta có đpcm
1) Cho a,b,c>0 tm a+b+c=3. Cmr \(\frac{1}{2+a^2+b^2}+\frac{1}{2+b^2+c^2}+\frac{1}{2+c^2+a^2}\le\frac{3}{4}\)
2) Cho a,b,c>0 tm \(a^2+b^2+c^2\le abc\).Cmr \(\frac{a}{a^2+bc}+\frac{b}{b^2+ca}+\frac{c}{c^2+ab}\le\frac{1}{2}\)
3) Cho a,b,c>0 tm \(\sqrt{a}+\sqrt{b}+\sqrt{c}=1\).Cmr \(\sqrt{\frac{ab}{a+b+2c}}+\sqrt{\frac{bc}{b+c+2a}}+\sqrt{\frac{ca}{c+a+2b}}\le\frac{1}{2}\)
Giúp mình mới nhé các bạn. Mình đang cần gấp
Cho a,b,c \(\ge\)0 TM \(a^2+b^2+c^2=1\) CM
\(\frac{c}{1+ab}+\frac{b}{1+ac}+\frac{a}{1+bc}\ge1\)
BĐT nhé ae: Với các ẩn dương nhé
1. abc=1. CM \(sigma\left(\frac{1}{2a^3+b^3+c^3+2}\right)\le\frac{1}{2}\)
2.\(a+b+c\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)CM \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
2/ GT <=> \(\left(a+b+c\right)abc\ge ab+bc+ca\)
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca}\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)abc}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
Sao hôm thứ 7 nghỉ
cho a , b , c >0. Chứng minh các bất đẳng thức :
1, ab + bc + ca \(\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
2, \(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)
3, \(ab+\frac{a}{b}+\frac{b}{a}\ge a+b+1\)
4, \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)
5, \(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
1.
Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)
\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
2.
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)
Cộng vế với vế:
\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)
3.
Từ câu b, thay \(c=1\) ta được:
\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)
4.
\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)
Dấu "=" xảy ra khi \(a=b=c\)
5.
\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)
Cộng vế với vế:
\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
1. bđt được viết lại thành
\(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)
Theo bđt AM-GM thì :
\(ab+bc\ge2\sqrt{ab\cdot bc}=2\sqrt{ab^2c}=2b\sqrt{ac}\)
Tương tự : \(bc+ca\ge2c\sqrt{ab}\); \(ab+ca\ge2a\sqrt{bc}\)
Cộng vế với vế
=> \(2\left(ab+bc+ca\right)\ge2\left(a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\right)\)
=> \(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)( đpcm )
Dấu "=" xảy ra <=> a=b=c
Cho: a,b,c > 0 và a + b + c = 3.
Chứng minh rằng:
a) \(\frac{a+b}{1+a}+\frac{b+c}{1+b}+\frac{c+a}{1+c}\ge ab+bc+ca\)
b) \(\frac{a}{ab+b^3}+\frac{b}{bc+c^3}+\frac{c}{ca+a^3}\ge\frac{3}{2}\)
Đề chơi căng nhỉ?
a) Dễ chứng minh VP =< 3
BĐT \(\Leftrightarrow\left(\frac{a+b}{1+a}-1\right)+\left(\frac{b+c}{1+b}-1\right)+\left(\frac{c+a}{1+c}-1\right)\ge0\)
\(\Leftrightarrow\frac{b-1}{1+a}+\frac{c-1}{1+b}+\frac{a-1}{1+c}\ge0\)
\(\Leftrightarrow\frac{\left(b-1\right)^2}{\left(1+a\right)\left(b-1\right)}+\frac{\left(c-1\right)^2}{\left(1+b\right)\left(c-1\right)}+\frac{\left(a-1\right)^2}{\left(1+c\right)\left(a-1\right)}\) >=0
Áp dụng BĐT Cauchy-Schwarz dạng Engel vào VT ta có đpcm.
P/s: Èo, sao đơn giản thế nhỉ? Em có làm sai chỗ nào chăng?
a, Ta có \(\frac{a+b}{a+1}=\frac{\left(a+b\right)\left(a+1\right)-a\left(a+b\right)}{a+1}=a+b-\frac{a\left(a+b\right)}{a+1}\)
Mà \(\frac{1}{a+1}\le\frac{a+1}{4a}\)
=> \(\frac{a+b}{1+a}\ge a+b-\frac{\left(a+1\right)\left(a+b\right)}{4}=\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}a^2-\frac{1}{4}ab\)
Khi đó
\(Vt\ge\frac{3}{2}\left(a+b+c\right)-\frac{1}{4}\left(a^2+b^2+c^2\right)-\frac{1}{4}\left(ab+bc+ac\right)\)
=> \(VT\ge\frac{9}{2}-\frac{1}{4}\left(9-2ab-2bc-2ac\right)-\frac{1}{4}\left(ab+bc+ac\right)\)
=> \(VT\ge\frac{9}{4}+\frac{1}{4}\left(ab+bc+ac\right)\)
Lại có \(ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=3\)
=> \(VT\ge ab+bc+ac\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c=1
b,Ta có \(\frac{a}{b\left(a+b^2\right)}=\frac{a+b^2-b^2}{b\left(a+b^2\right)}=\frac{1}{b}-\frac{b}{a+b^2}\)
Mà \(a+b^2\ge2b\sqrt{a}\)
=> \(\frac{a}{b\left(a+b^2\right)}\ge\frac{1}{b}-\frac{1}{2\sqrt{a}}\)
Lại có \(\frac{1}{\sqrt{a.1}}\le\frac{1}{2}\left(\frac{1}{a}+1\right)\)
=> \(\frac{a}{b\left(a+b^2\right)}\ge\frac{1}{b}-\frac{1}{4}.\left(\frac{1}{a}+1\right)\)
Khi đó
\(VT\ge\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)
Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3\)
=> \(VT\ge\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c=1
Bất đẳng thức được viết lại thành
\(\sum\frac{3-a}{1+a}\ge ab+bc+ca\)
Mà \(ab+bc+ca\le3\) nên ta chỉ cần chứng minh
\(\sum\frac{3-a}{1+a}\ge3\)
Ta chứng minh bất đẳng thức phụ sau
\(\frac{3-a}{1+a}\ge2-a\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)
Thiết lập các bất đẳng thức tương tự ta có điều phải chứng minh
Cho a,b,c>0, chứng minh:\(\frac{1}{a^2+ab+bc}+\frac{1}{b^2+bc+ca}+\frac{1}{c^2+ca+ab}\ge\frac{\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\)
Cho a, b, c là các số thực dương thỏa mãn điều kiện: \(ab+bc+ca\le3abc.\)
Chứng minh rằng : \(\sqrt{2}\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\ge\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+3\)