Tính:
A = $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}-\frac{b^2}{a+b}-\frac{c^2}{b+c}-\frac{a^2}{c+a}$
Cho a,b,c thỏa mãn a+b+c=0
Tính\(G=\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}\)
\(D=\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
a+b+c=0 <=> a+b=-c ; a+c=-b ; b+c=-a
\(\frac{1}{b^2+c^2-a^2}=\frac{1}{\left(b-a\right)\left(a+b\right)+c^2}=\frac{1}{\left(b-a\right)\left(-c\right)+c^2}=\frac{1}{c\left(a-b+c\right)}=\frac{1}{-2bc}\)
Tương tự: \(\frac{1}{c^2+a^2-b^2}=\frac{1}{-2ca};\frac{1}{a^2+b^2-c^2}=\frac{1}{-2ab}\)
=>\(G=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=\frac{0}{-2abc}=0\)
áp dụng cô si ta có:
+)\(\frac{a^5}{b^3}+\frac{a^3}{b}\ge\frac{2a^4}{b^2};\frac{b^5}{c^3}+\frac{b^3}{c}\ge\frac{2b^4}{c^2};\frac{c^5}{a^3}+\frac{c^3}{a}\ge\frac{2c^4}{a^2}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge2\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)-\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)\)
+)\(\frac{a^4}{b^2}+a^2\ge\frac{2a^3}{b};\frac{b^4}{c^2}+b^2\ge\frac{2b^3}{c};\frac{c^4}{a^2}+c^2\ge\frac{2C^3}{a}\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)-\left(a^2+b^2+c^2\right)\)
+)\(\frac{a^3}{b}+ab\ge2a^2;\frac{b^3}{c}+bc\ge2b^2;\frac{c^3}{a}+ca\ge2c^2\)
\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\left(a^2+b^2+c^2\right)+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}-a^2-b^2-c^2\right)\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)+\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}-\frac{a^3}{b}-\frac{b^3}{c}-\frac{c^3}{a}\right)\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)\)
Cho a, b, c > 0. CM:
a) \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\)
b) \(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}\le\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
c) \(\frac{a^2+b^2}{a^2-2ab+b^2}+\frac{b^2+c^2}{b^2-2bc+c^2}+\frac{c^2+a^2}{c^2-2ac+a^2}\ge\frac{5}{2}\)
(a, b, c đôi một khác nhau)
a) \(A=\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)
b) \(B=\frac{\frac{3a}{a+b}}{\frac{2a}{a^2-2ab+b^2}}\)
c) \(C=\frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-\frac{b}{a}}:\frac{\frac{a^2}{b^2}-\frac{b^2}{a^2}}{\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
a) \(A=\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)
\(=\frac{2bc+b^2+c^2-a^2}{2bc}.\frac{\frac{a+b+c}{b+c}}{\frac{b+c-a}{b+c}}.\frac{b^2+c^2-b^2+2bc-c^2}{a+b+c}\)
\(=\frac{\left(b+c+a\right)\left(b+c-a\right)}{2bc}.\frac{a+b+c}{b+c-a}.\frac{2bc}{a+b+c}\)
\(=a+b+c\)
b) \(B=\frac{\frac{3a}{a+b}}{\frac{2a}{a^2-2ab+b^2}}\)\(=\frac{3a}{a+b}.\frac{\left(a-b\right)^2}{2a}=\frac{3\left(a-b\right)^2}{2\left(a+b\right)}\)
c) \(C=\frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-\frac{b}{a}}:\frac{\frac{a^2}{b^2}-\frac{b^2}{a^2}}{\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
\(=\frac{\frac{a^2+b^2}{ab}}{\frac{a^2-b^2}{ab}}:\frac{\frac{a^4-b^4}{a^2b^2}}{\frac{\left(a+b\right)^2}{a^2b^2}}\)
\(=\frac{a^2+b^2}{a^2-b^2}.\frac{\left(a+b\right)^2}{a^4-b^4}\)
\(=\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)\left(a-b\right)\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)}\)
\(=\frac{1}{\left(a-b\right)^2}\)
Cho a , b , c thỏa mãn \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}=\frac{2017}{2018}\)
Tính giá trị \(p=\frac{a^2}{a+c}+\frac{b^2}{b+a}+\frac{c^2}{c+b}\)
Cho a, b, c > 0. CM:
a)\(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\le\frac{3}{4}\)
b)\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\)
c)\(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Làm được câu nào thì làm giúp mình câu đó nhé!
Lần sau đăng ít một thôi toàn bài dài :v, ko phải ko làm mà là ngại làm
a)Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{2a+b+c}=\frac{a}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{b}{a+2b+c}\le\frac{1}{4}\left(\frac{b}{a+b}+\frac{b}{b+c}\right);\frac{c}{a+b+2c}\le\frac{1}{4}\left(\frac{c}{a+c}+\frac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{4}\)
Xảy ra khi \(a=b=c\)
b)Đặt \(THANG=abc\left(a^2+bc\right)\left(b^2+ac\right)\left(c^2+ab\right)>0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{b+c}{a^2+bc}-\frac{c+a}{b^2+ac}-\frac{a+b}{a^2+ab}\)
\(=\frac{a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-b^4c^2a^2-c^4a^2b^2}{THANG}\)
\(=\frac{\left(a^2b^2-b^2c^2\right)^2+\left(b^2c^2-c^2a^2\right)+\left(c^2a^2-a^2b^2\right)^2}{2THANG}\ge0\) (Đúng)
Xảy ra khi \(a=b=c\)
c)Ta có:\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}\)
Và \(\frac{b^2}{c^2+a^2}-\frac{b}{c+a}=\frac{bc\left(b-c\right)+ab\left(b-a\right)}{\left(c+a\right)\left(c^2+a^2\right)}\)
\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{ac\left(c-a\right)+bc\left(c-b\right)}{\left(b+a\right)\left(b^2+a^2\right)}\)
Cộng theo vế 3 đăng thức trên ta có:
\(VT-VP=Σ\left[\frac{ab\left(a-b\right)}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{ab\left(a-b\right)}{\left(a+c\right)\left(a^2+c^2\right)}\right]\)
\(=\left(a^2+b^2+c^2+ab+bc+ca\right)\cdotΣ\frac{ab\left(a-b\right)^2}{\left(b+c\right)\left(c+a\right)\left(b^2+c^2\right)\left(c^2+a^2\right)}\ge0\)
2 bài cuối full quy đồng mệt thật :v
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a-\frac{a^2}{a+b}+b-\frac{b^2}{b+c}+c-\frac{c^2}{c+a}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\)
\(\Leftrightarrow a^2\left(a+b\right)\left(a+c\right)+b^2\left(b+a\right)\left(b+c\right)+c^2\left(c+a\right)\left(c+b\right)\ge a^2\left(a+c\right)\left(b+c\right)+b^2\left(b+a\right)\left(c+a\right)+c^2\left(c+b\right)\left(a+b\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge a^2c^2+a^2b^2+b^2c^2\left(lđ\right)\)
\(\Leftrightarrow\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\ge a+b+c\)
cho a,b,c thỏa mãn: \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{c^2}{a+b}+\frac{a^2}{b+c}+\frac{b^2}{c+a}\ge\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}\)
thì \(|a|=|b|=|c|\)
Chứng minh rằng nếu a,b,c thỏa mãn bất đẳng thức:
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}>\frac{a^2}{c+a}+\frac{b^2}{a+b}+\frac{c^2}{b+c}\) thì |a|=|b|=|c|
\(\frac{a^2b+bc^2-1}{ac\left(a+c\right)}+\frac{b^2c+ca^2-1}{ab\left(a+b\right)}+\frac{c^2a+ab^2-1}{bc\left(b+c\right)}\)
\(=\frac{a^2b^2+b^2c^2-b}{a+c}+\frac{b^2c^2+c^2a^2-c}{a+b}+\frac{c^2a^2+a^2b^2-a}{b+c}\)
\(=\frac{\frac{1}{a^2}-\frac{1}{ac}+\frac{1}{c^2}}{a+c}+\frac{\frac{1}{b^2}-\frac{1}{ab}+\frac{1}{a^2}}{a+b}+\frac{\frac{1}{c^2}-\frac{1}{bc}+\frac{1}{b^2}}{b+c}\ge\frac{1}{ac\left(a+c\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ab\left(b+a\right)}\)
\(=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)