I came to ask this because I am really stuck at this problem. I have tried everything from arithmetic mean, geometric mean and harmonic mean. Also, I have tried playing with the variables and such, but it got me to nowhere.

If $a+b+c=1$; $a,b,c$ nonnegative, calculate the minimum of $$\frac{4+3abc}{ab+bc+ac}$$ All I've got so far is: $$\frac{3abc}{ab+bc+ac} \le \frac{1}{3}$$ But this is obviously on the wrong side of the inequality. Also, I think that $$\frac{1}{ab+bc+ac}\ge3$$ But I haven't been able to prove it.

Playing with the most possible and obvious values, one could think that the answer is 37/3, but the excercise is about proving it. Any help and little hints are greatly apprecieated.

  • $\begingroup$ Yes, thanks, edited. $\endgroup$
    – chubakueno
    May 7 '13 at 2:42
  • $\begingroup$ Wolfram sure does say it's $\frac{37}{3}$. $\endgroup$
    – Maazul
    May 7 '13 at 2:49
  • $\begingroup$ Yes, putting 1/3 in a,b,c yields $\frac{37}{3}$. Unfortunanetly wolfram uses numerical methods and doesn´t give a proof of why it is $\frac{37}{3}$ $\endgroup$
    – chubakueno
    May 7 '13 at 2:52
  • $\begingroup$ If we divide numerator and denominator by (abc), we get $\frac{\frac{4}{abc}+3}{a+b+c} = \frac{4}{abc}+3 $ since a+b+c =1.. Will it help somehow... $\endgroup$
    – Sachin
    May 7 '13 at 2:57
  • $\begingroup$ Did you try replacing $a=1-(b+c)$ in $\frac{4+3abc}{ab+bc+ac}$ and then taking partial derivatives of the result with respect to $b$ and then $c$ and checking where they both vanish? Check if it gives you values $b=c=\frac{1}{3}$. $\endgroup$
    – Maazul
    May 7 '13 at 2:59

It seems you are right and the minimum is indeed 37/3. We shall use the standard techniques for proving the inequalities.

If we put $a=b=c=1/3$ then the obtain the upper bound $37/3$ for the minimum. It rests to show that holds

(1) $\frac{4+3abc}{ab+bc+ac}\ge 37/3.$

At first we homogenize [Lee, Ch 3] the left side


Expanding and simplifying, we reduce inequality (1) to the form

$$12(a^3+b^3+c^3)\ge (a^2b+ab^2+ab^2+ac^2+b^2c+bc^2)+30abc,$$

which should follow from Muirhead Theorem [Lee, Ch. 3.3].


[Lee] Hojoo Lee. Topics in Inequalities - Theorems and Techniques (February 25, 2006).

  • $\begingroup$ Thank you for your answer and pointing me to such a fantastic source of problems, proofs and theorems. $\endgroup$
    – chubakueno
    May 7 '13 at 3:32
  • $\begingroup$ May be somewhere in the net is a newer version of this book. $\endgroup$ May 8 '13 at 6:00

Substitute $a=1-(b+c)$ in $\displaystyle\frac{4+3abc}{ab+bc+ac}$ to get

$$\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c} \tag{1}$$

Differentiating $(1)$ with respect to $b$ gives

$$\frac{\partial}{\partial b}\left(\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c}\right) =\frac{(3c^3-3c^2+4)(2b-c+1)}{\left(b^2+b(c-1)+c(c-1)\right)^2}$$

Setting this equal to zero yields

$$2b-c+1=0 \tag{2}$$

Differentiating $(1)$ with respect to $c$ gives

$$\frac{\partial}{\partial c}\left(\frac{4+3(1-(b+c))bc}{(1-(b+c))b+bc+(1-(b+c))c}\right) =\frac{(3b^3-3b^2+4)(2c-b+1)}{\left(c^2+c(b-1)+b(b-1)\right)^2}$$

Setting this equal to zero yields

$$2c-b+1=0 \tag{3}$$

Solving for $(2)$ and $(3)$ yields


Plug these values in $(1)$ and you end up with


  • $\begingroup$ Thank you for the effort and pointing me to an introduction of differentiation, the topic is now answered and closed. $\endgroup$
    – chubakueno
    May 7 '13 at 3:34
  • $\begingroup$ @Maazul (1) Equality to the zero of partial derivatives in the point $x$ are sufficient condition of an extremum only when $x$ is an interior point of the function domain. For a complete proof you should separately consider also cases $b=0$, $c=0$, $b=1$ and $c=1$. (2) Usually we simplify an expression before we differentiate it. :-) $\endgroup$ May 8 '13 at 6:00


Denote $u = ab + bc + ca$, we have the following two results

(1) $0 \le 3 u \le 1$,


(2) $9abc \ge 4u - 1$.


$$ 3abc+4 = \frac{9abc+12}{3} \ge \frac{4u + 11}{3} \ge \frac{4u + 11\times3u}{3} = \frac{37}{3} u. $$ which is the desired result.

Lemma 1

To show (1), we expand $$(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$$ which gives $$(a + b + c)^2 - 3 (ab+bc+ca) \ge 0$$ which means $1 \ge 3u$. The $u \ge 0$ part is obvious for $a$, $b$, and $c$ are nonnegative.

Lemma 2

For (2), we can use Schur's inequality $$a^3+b^3+c^3+3abc \ge a^2(b+c)+b^2(c+a)+c^2(a+b).$$ Since $b + c = 1 - a$, etc, we get $$2(a^3+b^3+c^3)+3abc \ge a^2+b^2+c^2.$$

Using $$ a^2 + b^2 + c^2 = (a+b+c)^2 - 2 (ab+bc+ca), $$ and $$ a^3 + b^3 +c^3 - 3abc = (a + b + c)^3 - 3 (a+b+c) (a b + b c + c a). $$ we get $$ 2(1-3u)+9abc \ge 1-2u, $$ which is Lemma 2.


Generally, we can change the coefficients a bit, $$ \frac{ 9 p \, a b c + q + p r + 3 q r } { a b + b c + c a + r } \ge p + 3q, $$ and the problem is the special case of $p = \frac{1}{3}, q = 4, r = 0$.

This problem inspires another one, and the use of Schur's inequality is suggested by Dylan.


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