Cho a+b=c
Chứng minh rằng (\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\))2 = \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Cho các số thực dương $a,b,c$ thỏa mãn $a+b+c=1$. Chứng minh rằng $\frac{a}{2a+b^{2}}+\frac{b}{2b+c^{2}}+\frac{c}{2c+a^{2}}\leq \frac{1}{7}\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )$
Ta có: \(2a+b^2=2a\left(a+b+c\right)+b^2=b^2+2a^2+2ab+2ac\)
\(\ge4ab+2ac+a^2\)
\(\Rightarrow\frac{a}{2a+b^2}\le\frac{a}{4ab+2ac+a^2}=\frac{1}{4b+2c+a}\)
\(\le\frac{1}{49}.\frac{49}{4b+2c+a}=\frac{1}{49}.\frac{\left(4+2+1\right)^2}{4b+2c+a}\)
\(\le\frac{1}{49}\left(\frac{16}{4b}+\frac{4}{2c}+\frac{1}{a}\right)=\frac{1}{49}\left(\frac{4}{b}+\frac{2}{c}+\frac{1}{a}\right)\)
CMTT: \(\frac{b}{2b+c^2}\le\frac{1}{49}\left(\frac{4}{c}+\frac{2}{a}+\frac{1}{b}\right);\frac{c}{2c+a^2}\le\frac{1}{49}\left(\frac{4}{a}+\frac{2}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2}\le\frac{1}{7}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( đpcm )
a) Cho \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
Chứng minh rằng: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)
b) Cho a, b, c khác nhau đôi một. Chứng minh rằng:
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)
\(\frac{yz+xz+xy}{xyz}=0\)
yz + xz + xy = 0
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2\times\left(xy+xz+yz\right)=x^2+y^2+z^2+2\times0=x^2+y^2+z^2\left(\text{đ}pcm\right)\)
a) Từ giả thiết suy ra: xy + yz + zx = 0
Do đó:
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=x^2+y^2+z^2\)
b) Đặt \(\frac{1}{a-b}=x\); \(\frac{1}{b-c}=y\); \(\frac{1}{c-a}=z\)
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a-b+b-c+c-a=0\)
Theo câu a ta có: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)
Suy ra điều phải chứng minh
a)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{xy+yz+xz}{xyz}=0\)
\(\Rightarrow xy+yz+xz=0\)
\(x^2+y^2+z^2=\left(x+y+z\right)^2\)
\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2+2xy+2yz+2xz\)
\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
Do \(xy+yz+xz=0\)
\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2\) ( đpcm )
b)
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2\)
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}+\frac{2}{\left(a-b\right)\left(b-c\right)}+\frac{2}{\left(b-c\right)\left(c-a\right)}+\frac{2}{\left(a-b\right)\left(c-a\right)}\)
\(\Rightarrow\frac{2}{\left(a-b\right)\left(b-c\right)}+\frac{2}{\left(b-c\right)\left(c-a\right)}+\frac{2}{\left(a-b\right)\left(c-a\right)}=0\)
\(\Rightarrow2\left(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)}\right)=0\)
\(\Rightarrow\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)}=0\)
\(\Rightarrow\frac{\left(c-a\right)^2\left(b-c\right)\left(a-b\right)+\left(a-b\right)^2\left(b-c\right)\left(c-a\right)+\left(b-c\right)^2\left(a-b\right)\left(c-a\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)
\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)\left[\left(a-b\right)+\left(b-c\right)+\left(c-a\right)\right]}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)
\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)\left[\left(-a+a\right)+\left(-b+b\right)+\left(-c+c\right)\right]}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)
\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right).0}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)
\(\Rightarrow0=0\) ( đpcm )
1, cho a,b,c là các số thực dương chứng minh rằng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{2a+b}{a\left(a+2b\right)}+\frac{2b+c}{b\left(b+2c\right)}+\frac{2c+a}{c\left(a+2c\right)}\)
2,cho x,y,z thỏa mãn x+y+z=5 và xy+yz+xz=8 chứng minh rằng \(1\le x\le\frac{7}{3}\)
3, cho a,b,c>0 chứng minh rằng\(\frac{a^2}{2a^2+\left(b+c-a\right)^2}+\frac{b^2}{2b^2+\left(b+c-a\right)^2}+\frac{c^2}{2c^2+\left(b+a-c\right)^2}\le1\)
4,cho a,b,c là các số thực bất kỳ chứng minh rằng \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\left(ab+bc+ac-1\right)^2\)
5, cho a,b,c > 1 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)chứng minh rằng \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le\sqrt{a+b+c}\)
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
Cho a, b, c lớn hơn 0. Chứng minh rằng:
\(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\frac{a}{b^2}+\frac{1}{a}\ge\frac{2}{b}\) BĐT Cô-si
Tương tự suy ra đpcm
Cho a, b, c >0. Chứng minh rằng \(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Áp dụng bất đẳng thức Cô-si ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Cho các số dương a,b,c thỏa mãn a+b+c=1. Chứng minh rằng:\(\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Bạn tham khảo:
Câu hỏi của Phạm Minh anh - Toán lớp 9 | Học trực tuyến
Cho a, b, c là các số dương và a+b+c=1 chứng minh rằng: \(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{15}{4}\)
vì \(a+b+c=1\)
\(< =>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=3+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}+\frac{a}{c}\)
\(=3+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\)
ta có pt:
\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(3+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\right)\)
\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{3}{4}+\frac{a^2+b^2}{4ab}+\frac{b^2+c^2}{4bc}+\frac{c^2+a^2}{4ca}\)
áp dụng bđt cô- si( cauchy) gọi pt là P
\(P\ge2\sqrt{\frac{ab}{a^2+b^2}\frac{a^2+b^2}{4ab}}+2\sqrt{\frac{bc}{b^2+c^2}\frac{b^2+c^2}{4bc}}+2\sqrt{\frac{ca}{c^2+a^2}\frac{c^2+a^2}{4ca}}+\frac{3}{4}\)
\(P\ge2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+\frac{3}{4}\)
\(P\ge2.\frac{1}{2}+2.\frac{1}{2}+2.\frac{1}{2}+\frac{3}{4}\)
\(P\ge1+1+1+\frac{3}{4}=\frac{15}{4}\)
dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)
<=>ĐPCM
Cho a, b, c > 0. Chứng minh rằng: \(\frac{a+b}{a^2+bc}+\frac{b+c}{b^2+ac}+\frac{c+a}{c^2+ab}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Cho a;b;c>0 . Chứng minh rằng : \(\frac{a+b}{a^2+bc}+\frac{b+c}{b^2+ac}+\frac{c+a}{c^2+ab}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Ta có: \(\frac{a+b}{bc+a^2}+\frac{b+c}{ac+b^2}+\frac{c+a}{ab+c^2}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a+b}{bc+a^2}-\frac{b+c}{ac+b^2}-\frac{c+a}{ab+c^2}\ge0\)
\(\Leftrightarrow\frac{a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-b^4c^2a^2-c^4a^2b^2}{abc\left(bc+a^2\right)\left(ac+b^2\right)\left(ab+c^2\right)}\ge0\)
\(\Leftrightarrow\frac{2a^4b^4+2b^4c^4+2c^4a^4-2a^4b^2c^2-2b^4c^2a^2-2c^4a^2b^2}{2abc\left(bc+a^2\right)\left(ca+b^2\right)\left(ab+c^2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a^2b^2-b^2c^2\right)^2+\left(b^2c^2-c^2a^2\right)^2+\left(c^2a^2-a^2b^2\right)^2}{2abc\left(bc+a^2\right)\left(ac+b^2\right)\left(ab+c^2\right)}\ge0\)(Đúng) (do a, b, c>0 )
bạn ơi mik chỉ làm ngếu ngáo thôi nhé :)) đúng thì đúng mà sai thì thôi nhé :)) cách mình tự chế nhé
đặt \(\frac{a+b}{a^2+bc}+\frac{b+c}{b^2+ac}+\frac{c+a}{c^2+ab}=Pain\)
áp dụng định lí six paths of Pain :) ta có
\(\frac{\left(a+b\right)}{a^2+bc}=\frac{\left(a+b\right)}{\frac{\left(a+b\right)}{\left(a+c\right)}}=\frac{1}{\left(a+c\right)}\) ( định lí Six Paths of Pain ) hì hì
thay vào ta được :)
\(\frac{1}{a+c}+\frac{1}{b+a}+\frac{1}{c+b}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
áp dụng cô si sáp cho 2 số ta có
\(\frac{1}{a+c}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{c}\right)\) luôn đúng
\(\frac{1}{b+a}\le\frac{1}{2}\left(\frac{1}{b}+\frac{1}{a}\right)\) luôn đúng
\(\frac{1}{c+b}\le\frac{1}{2}\left(\frac{1}{c}+\frac{1}{b}\right)\) luôn đúng
cộng các vế lại ta được và rút 2/2 ta được :))
\(Pain\le\frac{1}{2}\left(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\right)=\frac{2}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
hình như BDT đã được chứng minh :))
theo bài của bạn Phạm quốc cường ta có :))
\(\frac{a+b}{a^2+bc}+\frac{b+c}{b^2+ac}+\frac{c+a}{c^2+ab}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) luôn đúng :))
tức là \(\frac{1}{a+c}+\frac{1}{b+a}+\frac{1}{c+b}=\frac{a+b}{a^2+bc}+\frac{b+c}{b^2+ac}+\frac{c+a}{c^2+ab}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)luôn đúng :))
tức là định Lí six paths of Pain luôn đúng :))
dấu = xảy ra khi nào thì mình éo biết được :))
: các thành phần trẩu tre éo làm thì đừng tích sai cho mình nhé :)) mik ms lớp 7 thôi còn gà lắm :))