## SUBJECT 2.

Determine the elements of the set

$$
S=\left\{(a, b, c, d) \left\lvert\, 2 a+3 b+5 c+7 d \leq 174-\frac{8}{a}-\frac{27}{b}-\frac{125}{c}-\frac{343}{d}\right., \text { with } a, b, c, d \in(0, \infty)\right\}
$$
## Subject 2.

$2 a+3 b+5 c+7 d \leq 174-\frac{8}{a}-\frac{27}{b}-\frac{125}{c}-\frac{343}{d} \Leftrightarrow$

$\Leftrightarrow\left(2 a+\frac{8}{a}\right)+\left(3 b+\frac{27}{b}\right)+\left(5 c+\frac{125}{c}\right)+\left(7 d+\frac{343}{d}\right) \leq 174$ (1).

$1 \mathbf{p}$

Given that $a, b, c, d \in(0, \infty)$, applying the inequality of means we get:

$2 a+\frac{8}{a} \geq 2 \sqrt{2 a \cdot \frac{8}{a}}=8$, with equality when $2 a=\frac{8}{a}, a^{2}=4, a=2$ $\qquad$
$3 b+\frac{27}{b} \geq 2 \sqrt{3 b \cdot \frac{27}{b}}=18$, with equality when $3 b=\frac{27}{b}, b^{2}=9, b=3 \ldots \ldots \ldots \ldots \ldots . . . .1 p$

$5 c+\frac{125}{c} \geq 2 \sqrt{5 c \cdot \frac{125}{c}}=50$, with equality when $5 c=\frac{125}{c}, c^{2}=25, c=5 \ldots \ldots \ldots \ldots .1 \mathbf{p}$

$7 d+\frac{343}{d} \geq 2 \sqrt{7 d \cdot \frac{343}{d}}=98$, with equality when $7 d=\frac{343}{d}, d^{2}=49, d=7 \ldots \ldots .1 \mathbf{p}$ By summing up we find that:

$\left(2 a+\frac{8}{a}\right)+\left(3 b+\frac{27}{b}\right)+\left(5 c+\frac{125}{c}\right)+\left(7 d+\frac{343}{d}\right) \geq 8+18+50+98=174$ (2).........1p

From (1) and (2) it follows that $\left(2 a+\frac{8}{a}\right)+\left(3 b+\frac{27}{b}\right)+\left(5 c+\frac{125}{c}\right)+\left(7 d+\frac{343}{d}\right)=174$, and equality is achieved for $(a, b, c, d)=(2,3,5,7)$ $1 p$