Nice-easy!

[dduclam]

Let a,b,c be nonnegative real numbers which sum 3. Find the maximum value of

[tex]P=a(b-c)^2+b(c-a)^2+c(a-b)^2[/tex]

[dduclam]

Hint: Maximum value of P is [tex]\frac{27}{4}[/tex]

[Math Tutor]

Could you write the proof, please?

[dduclam]

Could you write the proof, please?

Ok

Let a,b,c be nonnegative real numbers which sum 3. Find the maximum value of

[tex]P=a(b-c)^2+b(c-a)^2+c(a-b)^2[/tex]

WOLG assume [tex]a\ge b\ge c\ge 0[/tex]

[tex]=> P\le a(b+c)^2+ba^2+ca^2=a(b+c)(a+b+c)=3.a.(b+c)\le 3.(\frac{a+b+c}2)^2=\frac{27}4[/tex] (by AM-GM)

Max [tex]P=\frac{27}4[/tex] when [tex]a=b=\frac3{2},c=0[/tex] or any cyclic permutation