BAM k is max

We know that

$a+b+c\geq\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$ and the equality holds when $a=b=c$

$a+b+c\geq3\sqrt[3]{abc}$ and the equality holds when $a=b=c$

So, when $a=b=c$ , we have

$3\sqrt[3]{abc}=\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$

This lead to

$a+b+c+3\sqrt[3]{abc}\geq2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$

Hence, $k=2$

This is the OP's solution.

As with all such inequality problems, the first step is to guess a value of $k$ then prove it. Normally the maximum occurs when all the variables are equal, i.e. $a=b=c$ , and here we have $6 \geq 3k \leftrightarrow k \leq 2$ . Now we will prove that:

$a+b+c+3\sqrt[3]{abc} \geq 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})$

Now, motivated by the fact that we let $a = b = c$ , we can try to smooth the variables till they are equal. See here for more about the principle of smoothing. Basically, suppose WLOG that $a \leq b \leq c$ and also let $f(a,b,c) = a+b+c+3\sqrt[3]{abc} - 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})$ . Now, we use a change of variables, as follows:

$a_1 = a, b= b_1 = c_1 = \sqrt{bc} = j$

Then we will have:

$f(a_1,b_1,c_1) = (a+2c+3 \times a^{\frac{1}{3}} \times j^{\frac{2}{3}}) - 2[j + (2aj)^{\frac{1}{2}}]$

$= a + 3\times a^{\frac{1}{3}} \times j^{\frac{2}{3}} - 4(aj)^{\frac{1}{2}} \geq 0$

Where the equality holds iff $a = 0$ or $a =b = c$ .

By the smoothing principle, let us prove that $f(a,b,c) \geq f(a_1,b_1,c_1)$ . Remark first that

$f(a,b,c) - f(a_1,b_1,c_1) = b+c - 2a^{\frac{1}{2}}(b^{\frac{1}{2}} + c^{\frac{1}{2}} - 2j^{\frac{1}{2}}) - 2j$

Because we have: $a \leq j$ , $b^{\frac{1}{2}} + c^{\frac{1}{2}} \geq 2j^{\frac{1}{2}}$ , thus we can conclude that:

$f(a,b,c) - f(a_1,b_1,c_1) \geq b+c-2\sqrt{j}(\sqrt{b}+\sqrt{c} - 2\sqrt{j}) - 2j$

$= b+c - 2(\sqrt{b} + \sqrt{c})\sqrt{j} + 2j$

$= (\sqrt{b} - \sqrt{j})^2 + (\sqrt{c} - \sqrt{j})^2$

$\geq 0$

This basically implies that $f(a,b,c) ≥ 0$ as desired. ■

Lets take a=b=c. K will be 2.

Let $a$ , $b$ , and $c$ all be equal to 1. Thus for the equation to be set perfectly equal, $k=2$ .

Simplifying first the inequality by dividing both sides by 3: (a+b+c)/3 + cube root (abc) is greater than or equal to k(sqrt (ab) + sqrt (bc) + sqrt (ac))/3.

One can see that AM - GM Inequality is the best tool for this problem. This implies that

(a+b+c)/3 is greater than or equal to cube root(abc)

and

(sqrt (ab) + sqrt (bc) + sqrt (ac))/3 is greater than or equal to cube root (abc)

Multiplying the second applied AM - GM inequality both sides by k.

k(sqrt (ab) + sqrt (bc) + sqrt (ac))/3 is greater than or equal to k cube root (abc).

Using properties of inequality:

[(a+b+c) - k(sqrt (ab) + sqrt (bc) + sqrt (ac))]/3 + cube root (abc) is greater than or equal to 0. [(a+b+c) - k(sqrt (ab) + sqrt (bc) + sqrt (ac))]/3 is greater than or equal to - cube root (abc).

Remember from the applied AM - GM Inequality that: k(sqrt (ab) + sqrt (bc) + sqrt (ac))/3 is greater than or equal to k cube root (abc).

Hence, [(a+b+c) - k(sqrt (ab) + sqrt (bc) + sqrt (ac))]/3 is greater than or equal to (1-k) cube root (abc).

Comparing the yielded inequality to the the given has the same left-hand side but the right-hand side does not. This inequality satisfies when 1-k = -1. Hence, k = 2 and this is the maximum value that makes this inequality true.