Using rearrangement inequalities prove the following inequality:
Let $a,b,c$ be positive real numbers satisfying $abc=1$. Prove that
$ab^2+bc^2+ca^2 \geq a+b+c.$
Thanks :)
Using rearrangement inequalities prove the following inequality:
Let $a,b,c$ be positive real numbers satisfying $abc=1$. Prove that
$ab^2+bc^2+ca^2 \geq a+b+c.$
Thanks :)
For fun, let us do this one also with Lagrange multipliers:
In[33]:= f = a b^2 + b c^2 + c a^2 - a - b - c Out[33]= -a - b + a b^2 - c + a^2 c + b c^2 In[34]:= sol = Solve[ {D[f, a] == k b c, D[f, b] == k a c, D[f, c] == k a b, a b c == 1, a > 0, b > 0, c > 0}, {a, b, c, k} ] Out[34]= {{a -> 1, b -> 1, c -> 1, k -> 2}} In[35]:= f /. sol[[1]] Out[35]= 0
One has to check that my $f$ can only have a minimum, but that is easy.
To reiterate my comment above, Rearrangement Inequality needs some ordering in the variables, and hence cannot be applied here as the given inequality isn't symmetric wrt a,b,c.
To give a simple proof by AM-GM just note that $a^2c + a^2c + ab^2 \ge 3 \sqrt[3]{a^5b^2c^2} = 3a$ so adding the two other similar inequalities we get $ab^2 + bc^2 + ca^2 \ge a + b + c$.
I think I have the solution using arrangements inequalities.(source: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=497213)
We make the substitution $\displaystyle a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$. We have now:
$\frac{z^2}{xy}+\frac{x^2}{yz}+\frac{y^2}{xz} \geq \frac{y}{x}+\frac{z}{y}+\frac{x}{z}.$
So: $z^3+x^3+y^3\geq y^2z+z^2x+x^2y.$ And this inequality can be solved using rearrangements inequality.
Let $x \geq y \geq z$. Using rearrangement inequality for $(x^2,y^2,z^2)$ and $(x,y,z)$ we conclude that $x^2 \cdot x+ y^2 \cdot y + z^2 \cdot z \geq x^2 \cdot y+y^2 \cdot z+ z^2 \cdot x.$