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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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What have you tried? Is there a specific part of the problem you're having trouble with? – axblount Sep 5 '12 at 14:14
Note the Rearrangement Inequality cannot be applied here as you have to assume some ordering of $a,b,c$ in order to apply it, and the given inequality isn't symmetric wrt $a,b,c$.. – Rijul Saini Sep 5 '12 at 14:20
@axblount the inequality can be written like : $\frac{b}{c}+\frac{c}{a}+\frac{a}{b} \geq a+b+c$. and now I want to form a sequence with same monotony with $(a,b,c)$ where I supose that : $a\geq b\geq c$. I should want to use reaarangemnt inequalities for: $\left(\frac{b}{c}, \frac{c}{a}, \frac{a}{b}\right)$ and $(c,a,b)$. – Iuli Sep 5 '12 at 14:20
@RijulSaini I'm very sure it can be apply the rearrangement inequality. This inequality is from the chapter titled Rearrangement Inequalities. – Iuli Sep 5 '12 at 14:25
Do you know that if $\prod^n x_i = 1 $ then $\sum^n x_i \geq n$? – Pedro Tamaroff Sep 5 '12 at 18:31

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) – robjohn Sep 5 '12 at 16:01
up vote 4 down vote accepted

I think I have the solution using arrangements inequalities.(source:

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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No! Why can you assume an ordering on $x,y,z$ if there is no symmetry? – Rijul Saini Sep 5 '12 at 23:38
Iuli's proof is substantially correct and you can prove the second displayed equation using rearrangements. This method doesn't necessarily have to prove symmetric inequalities. You just have to take the correct permutation to get $y^2z+z^2x+x^2y$. For instance, supposing that $y\geq x\geq z$, the same inequality can be proved. – Francesco Sica Sep 6 '12 at 3:55
@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 $(x^2, y^2, z^2)$. – AlbertH Sep 9 '12 at 0:34

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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Thinking is overrated. – Mariano Suárez-Alvarez Sep 5 '12 at 20:54
Some people just don't like a challenge... – Zarrax Sep 5 '12 at 21:21

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