cho a,b,c
a+b=167
b+c=98
c+a=107
a=? b=? c=?
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LB
2
10 tháng 10 2021
1, Áp dụng BĐT cosi cho a,b,c>0
\(ab+bc\ge2\sqrt{ab^2c}=2b\sqrt{ac}\\ bc+ca\ge2\sqrt{abc^2}=2c\sqrt{ab}\\ ca+ab\ge2\sqrt{a^2bc}=2a\sqrt{bc}\)
Cộng VTV 3 BĐT trên:
\(\Leftrightarrow2\left(ab+bc+ac\right)\ge2\left(b\sqrt{ac}+a\sqrt{bc}+c\sqrt{ab}\right)\\ \Leftrightarrow ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)
10 tháng 10 2021
\(2,\)
Ta có
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\\ \Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Áp dụng BĐT cm ở câu 1
Suy ra đpcm
KK
0
QT
19 tháng 5 2018
\(\sum\dfrac{a}{b^2+bc+c^2}\ge\dfrac{\left(a+b+c\right)^2}{ab^2+abc+ac^2+bc^2+abc+ba^2+ca^2+abc+cb^2}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}=\dfrac{a+b+c}{ab+bc+ac}\)
L
0
19 tháng 11 2019
a)\(VT=\sum_{cyc}\frac{ab^3+ab^2c+a^2bc}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\frac{\sum_{cyc}\left(ab^3+ab^2c+a^2bc\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)\(\le\frac{\sum_{cyc}ab\left(a^2+b^2\right)+abc\left(a+b+c\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}=VP\)
HN
0
MK
1
7 tháng 7 2019
\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{a^2+ab+bc+ca}=\frac{a-bc}{\left(a+b\right)\left(c+a\right)}\)
\(=\left(a-bc\right)\sqrt{\frac{1}{\left(a+b\right)^2\left(c+a\right)^2}}\le\frac{\frac{a-bc}{\left(a+b\right)^2}+\frac{a-bc}{\left(c+a\right)^2}}{2}=\frac{a-bc}{2\left(a+b\right)^2}+\frac{a-bc}{2\left(c+a\right)^2}\)
Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)
=> \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{a-bc+b-ca}{2\left(a+b\right)^2}+\frac{b-ca+c-ab}{2\left(b+c\right)^2}+\frac{a-bc+c-ab}{2\left(c+a\right)^2}\)
\(\frac{\left(a+b\right)\left(1-c\right)}{2\left(a+b\right)\left(1-c\right)}+\frac{\left(b+c\right)\left(1-a\right)}{2\left(b+c\right)\left(1-a\right)}+\frac{\left(c+a\right)\left(1-b\right)}{2\left(c+a\right)\left(1-b\right)}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)