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\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}\)
\(\frac{b^2}{a^2+c^2}-\frac{b}{a+c}=\frac{ab\left(b-a\right)+bc\left(b-c\right)}{\left(a^2+c^2\right)\left(a+c\right)}\)
\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{ac\left(c-a\right)+bc\left(c-b\right)}{\left(a^2+b^2\right)\left(a+b\right)}\)
Cộng các vế ta có:
\(\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(=ab\left(a-b\right)\left[\frac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\frac{1}{\left(a^2+c^2\right)\left(a+c\right)}\right]\)\(+ac\left(a-c\right)\left[\frac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\frac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\)
\(+bc\left(b-c\right)\left[\frac{1}{\left(a^2+c^2\right)\left(a+c\right)+}-\frac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\)
Giả sử \(a\ge b\ge c>0\)thì
\(\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)>0\)
=> \(\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Dấu " = " xảy ra <=> a=b=c
Với a,b,c > 0
Áp dụng bđt cosi cho 2 số dương \(\frac{a^2}{b^2}\)và \(\frac{b^2}{c^2}\), ta có:
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}}=2\frac{a}{c}\) (1)
CMTT: \(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge2\frac{c}{b}\)(2)
\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\frac{b}{a}\)(3)
Từ (1), (2) và (3) cộng vế theo vế:
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{a^2}{b^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\frac{a}{c}+2\frac{c}{b}+2\frac{b}{a}\)
<=> \(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\)
<=> \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\)
Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\) (1)
Ta có : a+b+c khác 0
do nếu a+b+c=0=>\(\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}=1\)=>-3=1(Vô lí)
do a+b+c khác 0 nên ta nhân (a+b+c) vào (1)
=>\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
=>\(\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c\left(a+b\right)+c^2}{a+b}=a+b+c\)
=>\(\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)(ĐPCM)
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{ab}{c+a}+\frac{ac}{a+b}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{ac}{b+c}+\frac{bc}{c+a}=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{ab+bc}{c+a}+\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+a+b+c=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
cái này tương tự nà chỉ khác tử -> mẫu Câu hỏi của Thiên An - Toán lớp 9 - Học toán với OnlineMath
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Leftrightarrow\left(\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\right).\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{a}{\left(c-a\right)\left(b-c\right)}+\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a\left(c-a\right)+a.\left(a-b\right)+b.\left(a-b\right)+b.\left(b-c\right)+c.\left(b-c\right)+c.\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{ac-a^2+ab-ac+ba-b^2+b^2-bc+bc-c^2+c^2-ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+0=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
đpcm
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}=\frac{a^2+b^2+c^2}{2.\left(a+b+c\right)}\)
\(\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}=\frac{a^2+b^2+c^2}{2.\left(a+b+c\right)}\)
=> \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}=\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}\left(=\frac{a^2+b^2+c^2}{2.\left(a+b+c\right)}\right)\)