Minimum of 1/a + 4/b + 9/c When a + b + c = 6

For positive real numbers $a,b,c$ with $a+b+c=6$, the expression is minimized by assigning the variables in proportion to the square roots of their numerators. Here that gives $a:b:c=1:2:3$. The minimum value is 6, attained at $(a,b,c)=(1,2,3)$.

Apply Cauchy–Schwarz

The Engel form of the Cauchy–Schwarz inequality states that, for positive $x_i$ and real $u_i$,

$$\sum_i \frac{u_i^2}{x_i} \geq \frac{(\sum_i u_i)^2}{\sum_i x_i}.$$

Choose $(u_1,u_2,u_3)=(1,2,3)$ and $(x_1,x_2,x_3)=(a,b,c)$. Then

$$\frac{1}{a}+\frac{4}{b}+\frac{9}{c} \geq \frac{(1+2+3)^2}{a+b+c}.$$

Substitute the constraint $a+b+c=6$:

$$\frac{1}{a}+\frac{4}{b}+\frac{9}{c} \geq \frac{36}{6}=6.$$

Thus no allowed choice of $a,b,c$ can make the expression smaller than 6. This is a lower bound; to prove it is the actual minimum, check that equality can occur.

Find the equality case

In this form of Cauchy–Schwarz, equality holds when the ratios $u_i/x_i$ are all equal. In this problem, that condition is

$$\frac{1}{a}=\frac{2}{b}=\frac{3}{c}.$$

Equivalently, $a:b:c=1:2:3$. Write $a=k$, $b=2k$, and $c=3k$. The sum constraint gives

$$k+2k+3k=6,$$

so $k=1$ and therefore $(a,b,c)=(1,2,3)$. These are positive numbers and satisfy the constraint. Substitution confirms the bound is attained:

$$\frac{1}{1}+\frac{4}{2}+\frac{9}{3}=1+2+3=6.$$

Since the expression is always at least 6 and equals 6 for an allowed triple, its minimum is exactly 6.

Why the variables are not equal

The numerators are different, so splitting the available sum equally among $a,b,c$ does not generally minimize the expression. The equality condition shows how the allocation should change: a term with a larger numerator needs a proportionally larger denominator. In general, for positive $p_i$ and positive $x_i$ with fixed sum $S$, Cauchy–Schwarz gives

$$\sum_i \frac{p_i}{x_i} =\sum_i \frac{(\sqrt{p_i})^2}{x_i} \geq \frac{(\sum_i \sqrt{p_i})^2}{S}.$$

Equality occurs when $x_i$ is proportional to $\sqrt{p_i}$. In this question, the numerators are $1,4,9$, whose square roots are $1,2,3$. Their sum is 6, so the equality allocation under a total of 6 is exactly $(1,2,3)$.

Common mistakes

• Using $a=b=c=2$ without checking equality. This is a valid triple, but it gives $1/2+4/2+9/2=7$, not the minimum.
• Treating the numerators themselves as the allocation ratio. The equality condition uses their square roots: $a:b:c=\sqrt{1}:\sqrt{4}:\sqrt{9}=1:2:3$.
• Stopping at a lower bound. An inequality alone shows the expression is at least 6. You must also exhibit an allowed triple where it equals 6 to establish the minimum.
• Allowing a variable to be zero. The expression contains reciprocals, so the variables must be strictly positive. The equality triple satisfies this requirement.