IMO Math

 Let $n > 1$ be an integer and let $f(x) = x^n + 5 \cdot x^{n-1} + 3.$ Prove that there do not exist polynomials $g(x),h(x),$ each having integer coefficients and degree at least one, such that $f(x) = g(x) \cdot h(x).$ Solution 1 (overkill):  Since $5>1+3$, from Perron's Criterion this polynomial is irreducible over the integers as desired. Solution 2 (normal solution): Assume FTSOC that there do exist $g(x), h(x)$ such that $f(x)=g(x)\cdot h(x)$. Observe that there are no integer roots of $f(x)$ from the Rational Root Theorem. Thus, $g(x), h(x)$ cannot be linear, and their degree is greater than or equal to $2$. We have $g(0)h(0)=3$. WLOG $g(0)=1$. Let $r_1, r_2, \dots r_j \in\mathbb{C}$ be the roots of $g(x)$. We have $r_1r_2\dots r_n=\pm 1$. Multiplying the equalities $r_i^{n-1}(r_i+5)=-3$ for all $1\leq i\leq j$, we obtain\[\vert g(-5)\vert = \vert (r_1+5)(r_2+5)\dots (r_m+5)\vert = 3^m.\]But $g(-5)f(-5)=3$, contradiction.

 Prove for all positive reals $a,b,c$, $$\sum_{cyc} \frac{a}{\sqrt{3ab+bc}} \ge \frac 32.$$ I'll wait for comments and will post the solution in 2 days! EDIT - Well now that it's been 3 days (oops i forgot to post), I'll show my solution now.  By Holder,$$\left( \sum_{cyc} \frac{a}{\sqrt{3ab+bc}} \right)^2 \left( \sum_{cyc} a(3ab+bc) \right) \ge (a+b+c)^3$$ So it suffices to show$$(a+b+c)^3 \ge \frac 94 \sum_{cyc} a(3ab+bc)$$ Expanding and cancelling terms, we wish to show$$f(a,b,c)=4\sum_{cyc} a^3 + 12\sum_{cyc} b^2a-15\sum_{cyc} a^2b-3abc\ge 0$$ Claim: $f(a,b,c) \le f(a+d, b+d, c+d)$ if $d>0$ Proof:$$f(a+d,b+d,c+d)-f(a,b,c) =d (4\sum_{cyc} 3a^2 + 12\sum_{cyc} (b^2+2ba) - 15\sum_{cyc} (a^2+2ba) - 3(ab+bc+ca)) $$ $$+d^2 (4\sum_{cyc} 3a + 12 \sum_{cyc} (2b+a) - 15\sum_{cyc} (2a+b) - 3(a+b+c)$$ The $d^2, d^3$ stuff cancel. So $f(a+d,b+d,c+d) - f(a,b,c) = \frac{9d}{2} \sum_{cyc} (a-b)^2 > 0$. Let $a=\min\{a,b,c\}$, then $f(a,a+b,a+c) \ge f(0,b,c)$. Let $x=\frac bc$, then

Last one. I promise.   Prove that for all positive real numbers $a,b,c$,\[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1.  \]

Since the last one was super easy, I have to do another one... right?  We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a + b$ on both sheets. Prove that after $m2^{m -1}$ steps, the sum of the numbers on all the sheets is at least $4^m$ .

 Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.

Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

 Find the smallest positive integer $n$ or show no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers $(a_1, a_2, \ldots, a_n)$ such that both $$a_1+a_2+\dots +a_n \quad \text{and} \quad \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}$$are integers.