ABCs to the Max

In the inequality, we let $a=b=c=1$ to bound the possible values of $N.$ This gives $4 \ge N.$

We now prove that $N = 4$ works; by AM-GM, we have

$\dfrac{a^2b^2c^2+ab+bc+ca}{4} \ge \sqrt[4]{(a^2b^2c^2)(ab)(bc)(ca)} = abc,$

so $a^2b^2c^2+ab+bc+ca \ge 4abc$ for all positive $a,b,c.$ Since $4 \ge N$ and $N = 4$ is possible, the maximum is $N = \boxed{4}.$

By the AM-GM inequality, we have $\frac{a^2b^2c^2+ab+bc+ca}{4} \ge \sqrt[4]{a^2b^2c^2\cdot ab\cdot bc\cdot ca}$ $=abc.$ Thus, $\boxed{N=4}$ .

Using the AM-GM inequality, we have ${ab+bc+ca \over 3} \ge \sqrt[3]{a^2b^2c^3}$ . So, we have:

$a^2b^2c^2+ab+bc+ca$

$\ge a^2b^2c^2+3\sqrt[3]{a^2b^2c^2} .... (1)$ .

So if $N$ satisfies $1$ , then it also satisfies the original inequality. If $N$ satisfies $1$ , then,

$a^2b^2c^2 + 3\sqrt[3]{a^2b^2c^2} \ge Nabc$ .

$\equiv p^4-Np+3 \ge 0$ where $p = (abc)^{1 \over 3}$ . Let $f(x) = x^4-Nx+3$ . This function has a single real minima at $x^\star=\left(N \over 4\right)^{1 \over 3}$ . If the value of $N$ is such that $f(x^\star) \ge 0$ , then $(1)$ is true for all $N$ . Substituting $x^\star$ in $f(x)=0$ , and solving for $N$ , we get $N=4$ . It is easy to verify that the inequality $a^2b^2c^2+ab+bc+ca \ge Nabc$ is not satisfied for $N=5$ by letting $a=b=c=1$ and noting that $1+3 \ngeq 5$ . So $N=4$ is the largest integer that satisfies the inequality.

From AM-GM inequality

We know that AM $\geq$ GM Taking $a^2b^2c^2, ab, bc, ca$ as four terms we get

$\frac{a^2b^2c^2+ab+bc+ca}{4} \geq \sqrt[4]{a^2b^2c^2.ab.bc.ca}$

$\Rightarrow \frac{a^2b^2c^2+ab+bc+ca}{4} \geq abc$

$\Rightarrow a^2b^2c^2+ab+bc+ca \geq 4abc$

Thus $N= \boxed{4}$

Use AM> =GM

Using directly AM-GM inequality

$\frac{a^2b^2c^2 + ab + bc + ca}{4} \geq (a^4b^4c^4)^\frac{1}{4} = abc$

Therefore, $a^2b^2c^2 + ab + bc + ca \geq 4abc$

Hence $N = 4$

Using $A.M \ge G.M$ inequality:

$(a^{2}b^{2}c^{2}+ab+bc+ca)/4 \ge (a^4b^4c^4)^{1/4}$ or $(a^{2}b^{2}c^{2}+ab+bc+ca) \ge 4abc$

Comparing N = 4

Consider a=1, b=1, c=1.
1+1+1+1≥N.
N=4

By AM-GM Inequality,

$a^2b^2c^2+ab+bc+ca$

$\geq 4\sqrt[4]{\strut (a^2b^2c^2)(ab)(bc)(ca)}$

$\geq4abc$

Hence, the maximum value of $N$ is 4.

This can be done easily by using the AM-GM inequality

it is clear enough that if one puts any value less than 1 like a=0.2,b=0.5,c=0.9 etc......... n is bound to exceed 10. similar is the case with nos. greater than 1 where the value of n goes on increasing with succeeding inputs........ but when we put a=1,b=1,c=1 we deduce n=4........which happens to be the largest integer at which this expression holds good for all positive nos

We see that a, b and c are squared on the LHS which furthermore only contains terms increasing the value of the LHS. We see that a, b and c are not squared but linearly multiplied with each other on the RHS. This and the fact that a, b and c are positive integers means that we just need to minimize the values of a, b and c. This means we simply need to plug in $(a,b,c) = (1,1,1)$ to arrive at the largest integer N. $N = (1^{2})(1^{2})(1^{2}) + (1)(1) + (1)(1) + (1)(1) = \boxed{4}$

O menor valor que as incógnitas podem assumir é 1, daí,

$1^2 \cdot 1^2 \cdot 1^2 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 \geq N \cdot 1 \cdot 1 \cdot 1 \\\\ 1 + 1 + 1 + 1 \geq N \\\\ 4 \geq N \\\\ N \leq 4$

Logo, $\boxed{N = 4}$

By AMGM, note

$ab + bc >= 2b\sqrt{ac}$

$ab + ac >= 2a\sqrt{bc}$

$bc + ac >= 2c\sqrt{ab}$

adding and dividing by 2 gives $ab + bc + ac >= b\sqrt{ac} + a\sqrt{bc} + c\sqrt{ab}$ substituting into the original equation and dividing by $abc$ $abc + 1/\sqrt{ac} + 1/\sqrt{bc} + 1/\sqrt{ab} >= N$ which clearly has a minimum when a = b = c = 1 hence N = 4

By AM-GM $$a^2b^2c^2 + ab + bc + ca \geq 4\sqrt[4]{a^4b^4c^4} = 4abc$$ On the other hand for $(a,b,c) = (1,1,1,1)$ we get $$4 \geq N$$ Hence $$N = \boxed{4}$$

a^2 b^2 c^2 +ab+bc+ca >= Nabc.......Dividing each side of inequality by abc....................... As all three numbers a,b,c are positive, sign of inequality remains unchanged. Hence, abc+1/a+1/b+1/c>=N....Hence the greatest value of N is the least value of abc+1/a+1/b+1/c. Applyng AM- GM inequality, we get that abc+1/a+1/b+1/c>=4.......Hence , N max =4

$\frac{a^{2}*b^{2}*c^{2}+ab+bc+ca}{abc} = abc+a+b+c$

$abc+a+b+c>=N$

Minimum condition for equation to be true: a,b,c = 1. $1*1*1+1+1+1 = 4$ Therefore, largest integer * $N$ = 4 *

On dividing with abc both sides we get::

abc + (1/a) + (1/b) + (1/c) >= N

Symmetrically a = b = c = 1 should be the solution

ANS :-: 4

the largest number $N$ for all positive numbers $a, b, c$ , this mean $a = b = c = 1$ is included and it's the smallest possible solution for all $(a, b, c)$ then:

$a^2b^2c^2 + ab + bc + ca = (1^2)\times(1^2)\times(1^2) + (1)\times(1) + (1)\times(1) + (1)\times(1) = 4$

then $4 \geq N$ so $N = 4$ is solution.