giúp mình làm bài này đi rrooiif mình giúp cho
cho tam giac abc . co canh bc=12cm, duong cao ah=8cm
a> tinh s tam giac abc
b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách
c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a-\frac{a^2}{a+b}+b-\frac{b^2}{b+c}+c-\frac{c^2}{c+a}\ge a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\)
\(\Leftrightarrow a^2\left(a+b\right)\left(a+c\right)+b^2\left(b+a\right)\left(b+c\right)+c^2\left(c+a\right)\left(c+b\right)\ge a^2\left(a+c\right)\left(b+c\right)+b^2\left(b+a\right)\left(c+a\right)+c^2\left(c+b\right)\left(a+b\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge a^2c^2+a^2b^2+b^2c^2\left(lđ\right)\)
\(\Leftrightarrow\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\ge a+b+c\)
cho a, b, c > 0 cmr a^2/(b^2+c^2) + b^2/(c^2+a^2) + c^2/(a^2+b^2) >= a/(b+c) + b/(c+a) + c/(a+b)
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 \(|a|=|b|=|c|\)
a^2/(b-c)^2 + b^2/(a-c)^2 + c^2/(b-a)^2>=2
(a+b/a-b)^2+(b+c/b-c)^2+(c+a/c-a)^2>=2
Cho a, b, c # 0 thỏa mãn: a^2/b^2 + b^2/c^2 + c^2/a^2 = a/c + c/b + b/a và a+b+c=1. Tìm a, b, c
Cho a, b, c>0. Chứng minh:
a) a(b^2+bc+c^2)+b(c^2+ca+a^2)+c(a^2+ab+b^2)=<(1/3).(a+b+c)^3
b) a^3/(b^2+bc+c^2)+b^3/(a^2+ca+c^2)+c^3/(a^2=ab+b^2)>=(a+b+c)/3
cho a,b,c>0 cmr (a+b)^2/(a+b-c) + (b+c)^2/(b+c-a) + (c+a)^2/(a-b+c) >=4(a+b+c)
Cho a,b,c là độ dài ba cạnh tam giác.Chứng minh rằng:`sqrt{(a(b+c))/(a^2+(b+c)^2)}+sqrt{(b(c+a))/(b^2+(c+a)^2)}+sqrt{(c(a+b))/(c^2+(a+b)^2)}>sqrt2`
a/(b+c)+b/(c+a)+c/(a+b)=1 cmr (a^2)/(b+c)+(b^2)/(c+a)+(c^2)/(a+b)=0
