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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 :)

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    @RijulSaini Even if the end result isn't symmetric in $a$,$b$, and $c$, one might still apply the rearrangement inequality to a symmetric situation and end out with an asymmetric result.. the permutation can break the symmetry.2012-09-05

3 Answers 3

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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.

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    Some people just don't like a challenge...2012-09-05
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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 $ab^2 + bc^2 + ca^2 \ge a + b + c$.

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    Hit it hard enough with AM-GM and it falls down :-) (+1)2012-09-05
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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 $x^2 \cdot x+ y^2 \cdot y + z^2 \cdot z \geq x^2 \cdot y+y^2 \cdot z+ z^2 \cdot x.$

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    @RijulSaini To prove by rearrangement, an inequality of the type $a_1 b_1 + \dotsb + a_n b_n \geq a_1 b_{\pi(1)} + \dotsb + a_n b_{\pi(n)}$, where $\pi$ is a permutation of $\{1, \dotsc, n\}$, you only need that $(a_1, \dotsc, a_n)$ and $(b_1, \dotsc, b_n)$ does have the same ordering, that is $a_i \leq a_j \iff b_i \leq b_j$. This is certainly the case for $(x, y, z)$ and $(x^2, y^2, z^2)$.2012-09-09