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 :)

  • 2
    $\begingroup$ What have you tried? Is there a specific part of the problem you're having trouble with? $\endgroup$
    – axblount
    Commented Sep 5, 2012 at 14:14
  • $\begingroup$ 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 $a,b,c$.. $\endgroup$ Commented Sep 5, 2012 at 14:20
  • $\begingroup$ @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 $(c,a,b)$. $\endgroup$
    – Iuli
    Commented Sep 5, 2012 at 14:20
  • $\begingroup$ @RijulSaini I'm very sure it can be apply the rearrangement inequality. This inequality is from the chapter titled Rearrangement Inequalities. $\endgroup$
    – Iuli
    Commented Sep 5, 2012 at 14:25
  • $\begingroup$ Do you know that if $\prod^n x_i = 1 $ then $\sum^n x_i \geq n$? $\endgroup$
    – Pedro
    Commented Sep 5, 2012 at 18:31

4 Answers 4


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 $ab^2 + bc^2 + ca^2 \ge a + b + c$.

  • 1
    $\begingroup$ Hit it hard enough with AM-GM and it falls down :-) (+1) $\endgroup$
    – robjohn
    Commented Sep 5, 2012 at 16:01
  • $\begingroup$ Why the argument of Francesco Sica should not work? $\endgroup$
    – Thomas
    Commented Dec 2, 2021 at 4:11

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 $$x^2 \cdot x+ y^2 \cdot y + z^2 \cdot z \geq x^2 \cdot y+y^2 \cdot z+ z^2 \cdot x.$$

  • $\begingroup$ No! Why can you assume an ordering on $x,y,z$ if there is no symmetry? $\endgroup$ Commented Sep 5, 2012 at 23:38
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Sep 6, 2012 at 3:55
  • $\begingroup$ @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 $(x^2, y^2, z^2)$. $\endgroup$
    – AlbertH
    Commented Sep 9, 2012 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.

  • 2
    $\begingroup$ Thinking is overrated. $\endgroup$ Commented Sep 5, 2012 at 20:54
  • $\begingroup$ Some people just don't like a challenge... $\endgroup$
    – Zarrax
    Commented Sep 5, 2012 at 21:21

Just by Hölder inequality;


If $(\frac1a+\frac1b+\frac1c)\le(a+b+c)$ we're done. Otherwise;


since $(a+b+c)<(\frac1a+\frac1b+\frac1c)$ we're done again.


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