39thanks carl...but this is an olympiad ques??? surely not..
prove that
bc(b+c)+ca(c+a)ab(a+b)<=2(a3+b3+c3)
there r many sol to this sum but 1 ol just takes 2 steps lets see who get it
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7 Answers
39Inequality ka LHS kamsekam dhang se toh likh lo..
a2 + b2 ≥ 2ab [AM-GM]
=> a2 + b2 - ab ≥ ab
=> (a + b)(a2 - ab + b2) ≥ ab(a+b)
=> a3 + b3 ≥ ab(a + b)
Similarly, b3 + c3 ≥ bc(b + c), a3 + c3 ≥ ac(a + c)
Add the three inequalities to get your answer. Don't know if this is a short method, but it's the easiest I know.
39
1applying Muirhead's inequality we have
a^2b+b^2c+c^2a\leq a^3b^0+b^3c^0+c^3a^0
and
ab^2+bc^2+ca^2\leq a^3b^0+b^3c^0+c^3a^0
adding we get the result.....
so i used only two steps[4]........isse chota method dikhao.....
39So mine is street smart method, carl's is classicist's method, rick's is mathematician's method :P:P
1yeah thats it the reaarangement inequality method is the simplest 1
341By CS:
\frac{a^2}{b}+\frac{b^2}{a} \ge \frac{(a+b)^2}{a+b} = (a+b) \Rightarrow a^3 + b^3 \ge ab(a+b)