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A = \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
= \(a.\frac{a}{b+c}+b.\frac{b}{a+c}+c.\frac{c}{a+b}\)
=\(a.\frac{a}{b+c}+1-1+b.\frac{b}{a+c}+1-1+c.\frac{c}{a+b}+1-1\)
= \(\frac{a\left(a+b+c\right)}{b+c}-a+\frac{b\left(a+b+c\right)}{a+b}-b+\frac{c\left(a+b+c\right)}{a+b}-c\)
= \(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\)
= (a+b+c) - (a+b+c) = 0
1) Theo bđt AM-GM,ta có: \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}=a\)
Suy ra \(\frac{a^2}{b+c}\ge a-\frac{b+c}{4}\)
Thiết lập hai BĐT còn lại tương tự và cộng theo vế ta có đpcm
"Chấm" nhẹ hóng cao nhân ạ :)
P/s: mong các bác giải theo cách lớp 8 ạ :) Tặng 5SP / 1 câu nhé ;)
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Leftrightarrow\left(\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\right).\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{a}{\left(c-a\right)\left(b-c\right)}+\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a\left(c-a\right)+a.\left(a-b\right)+b.\left(a-b\right)+b.\left(b-c\right)+c.\left(b-c\right)+c.\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{ac-a^2+ab-ac+ba-b^2+b^2-bc+bc-c^2+c^2-ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+0=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
đpcm
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)
Có :a^2/b+c + b^2/c+a + c^2/a+b
= a.(a/b+c) + b.(b/c+a) + c.(c/a+b)
= a.(a/b+c + 1 - 1) + b.(b/c+a + 1 - 1) + c.(c/a+b + 1 - 1)
= a. a+b+c/b+c + b. a+b+c/c+a + c. a+b+c/a+b - (a+b+c)
= (a+b+c).(a/b+c + b/c+a + c/a+b) - (a+b+c)
= (a+b+c)-(a+b+c)
= 0
=> ĐPCM
Tk mk nha