# Multiple equality conditions

Prove that if $a, b, c$ are non-negative real numbers such that $a + b + c = 3$, then we have

$ab^2 + bc^2 + ca^2 + abc \leq 4 .$

Note by Calvin Lin
6 years ago

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Let $f(a,b,c)=ab^2+bc^2+ca^2+abc + k(a+b+c-3)$ (say this as equation $(1)$

Solving these four equations $\frac{\partial f}{\partial a} = 0$ $\frac{\partial f}{\partial b} = 0$ $\frac{\partial f}{\partial c} = 0$ $a+b+c=3$

we get $a=b=c=1 , k=-4$

substituting this back in equation $1$ , we get $f_{max}(a,b,c)=1+1+1+1-4(0) = 4$

$\textbf{Q.E.D}$

@Calvin Lin sir

- 5 years, 8 months ago

This demonstrates the point of this problem.

You did not perform the Lagrangian properly. At the IMO, this solution will be scored 0/7.
1. You did not state the equations.
2. You did not state how to solve the equations.
3. You missed out the equality case of $(1, 2, 0)$.

Staff - 5 years, 8 months ago

I didn't miss the case 1,2,0 doesnt satisfy the third equation ,i.e. , df/dc = 0

- 5 years, 8 months ago

You missed it because you didn't take care of the boundary condition. IE You didn't perform the Lagrangian properly.

E.g. What is the maximum of $f(x) = x^2$ on the interval $[-2, 2]$? Do you say that " $f' = 0 \Rightarrow x = 0$ hence the maximum is $f(0) = 0$? No, we still have to check the boundary points, where they need not satisfy $f' = 0$ in order to be a maximum on the restricted domain.

Staff - 5 years, 8 months ago

I know what you are trying to say . But what i know is that the min / max will be obtained using lagrangian even if we missed out boundary condition or other ordered pairs of equality

- 5 years, 8 months ago

No it will not.

For example, if you ignore the boundary condition when calculating the max of $f( x) = x^2$ on the closed interval $[1, -1]$, then you would not get any answer.

Staff - 5 years, 8 months ago

Okay ,, ya agree ! in this question we have to check at boundary.

ok i will keep in mind always to check at boundary

- 5 years, 8 months ago

This problem becomes quite a standard exercise when you use Lagrange multipliers. I think that the challenge is to prove the given inequality without using that which would be the reason for the problem being tagged under Algebra.

- 5 years, 8 months ago

i dont think so ... :)

- 5 years, 8 months ago

What Prasun said is actually true. Lagrange Multiplier makes this problem too easy. It's like you're using a chainsaw to cut a carrot.

On the other hand, you need to show that the extremal point you've found is a global maximum point as opposed to an inflection point or a global minimum point.

- 5 years, 8 months ago

This is quite a famous inequality, and there are several approaches that could be taken. Here is the simplest proof that I know of.

Define $f( a, b, c) = ab^2 + bc^2 + ca^2 + abc$. WLOG, we may assume that $a$ is the median of the 3, IE we either have $b \leq a \leq c$ or $b \geq a \geq c$.

Step 1: We will show that $f(a,b,c) \leq f ( a, b+c, 0 )$. This approach is known as Smoothing.

This follows by expanding both sides, and we want to compare $ab^2 + bc^2 + ca^2 + abc \leq a(b+c)^2$, which simplifies to

$bc^2 + ca^2 \leq abc + ac^2$

This is equivalent to

$c ( a - b ) ( a - c) \leq 0$

From the assumption that $a$ is the middle number, we know that $a-b, a-c$ will have different signs (or be 0). Since $c$ is non-negative, hence the entire product will be $\leq 0$.

This is essentially how we approach @Krishna Sharma 's Case 1.

Step 2: We show that subject to $a + b = 3$, we have $f(a, b, 0) \leq 4 = f ( 1, 2, 0)$.

By AM-GM, we get

$f(a, b, 0) = ab^2 = 4 \times a \times \frac{b}{2} \times \frac{b}{2} \leq 4 \left( \frac{a + \frac{b}{2} + \frac{b}{2} } { 3} \right) ^3 = 4 .$

Hence, the result follows.

However, I do not know of an easy way to motivate the approach, and in particular step 1. Any thoughts or comments?

Staff - 6 years ago

Sir, What is meant by IE ?

- 5 years, 6 months ago

It's just another way to write i.e. which means "that is".

- 5 years, 6 months ago

Minor typo in the last step where you use AM-GM.

$4\times a\times \left(\frac{b}{2}\right)^2\leq 4\left(\frac{a+\frac{b}{2}+\frac{b}{2}}{\color{#D61F06}{3}}\right)^3=4$

- 6 years ago

Thanks fixed.

Staff - 6 years ago

Why not assume a=b= c=1 and do it

- 6 years ago

- 6 years ago

This is just a comment, not a solution.

I remember seeing a stronger version of this inequality somewhere recently (probably on Math SE) which stated the following:

If $a,b,c$ are non-negative reals, then the following inequality holds:

$~\\~ab^2+bc^2+ca^2+abc\leq \frac{4}{27}(a+b+c)^3$

Our required inequality trivially follows from the stated inequality. However, proving the said stronger inequality seems to be harder than I expected. Let me see if I can think of a proof to that. For the time being, others are welcome to post their proof (if any) for the stated stronger inequality.

- 6 years ago

Note that the inequalities are equivalent. What happened was that we normalized the inequality, meaning that we made all of the terms have the same polynomial degree. To do so, we multiplied, where necessary, by $a + b + c = 3$.

This is a standard approach, and one that I would often (though not always) recommend to use if the terms have different degrees.

The weaker version that you saw, was most likely

$ab^2 + bc^2 + ca^2 \leq 4 \Leftrightarrow ab^2 + bc^2 + ca^2 \leq \frac{4}{27} ( a + b + c ) ^ 3 .$

Staff - 6 years ago

Is some info missing? Because substituting $a=b=c = \frac{4}{3}$ yields a value around 9.5

- 6 years ago

Ooops, the condition should have been $a+b+ c = 3$.

Staff - 6 years ago