I would appreciate if somebody could help me with the following problem.
For any two natural numbers $x$,$y$, find the maximum value of $A$ satisfying the following inequality $$3x^2+y^2+1\ge A(x^2+xy+x)$$ The above issue is the second round of the Korean secondary competition
My work : I tried to transform it into a quadratic equation for $x$ and use the discriminant to find it, but it is too difficult to interpret the discriminant under the condition that it is a natural number.
Your idea works very well!
Indeed, let $x=y=1$.
Thus, $A\leq\frac{5}{3}.$
We'll prove that $\frac{5}{3}$ is a maximal value.
Let $x=1+a$ and $y=1+b$,where $a$ and $b$ are integers and $a\geq0$, $b\geq0.$
Thus, we need to prove that $$3b^2-(5a-1)b+4a^2-2a\geq0,$$ which is obvious for $a=0$.
Let $a\geq1$.
Thus, we need to prove that: $$(5a-1)^2-12(4a^2-2a)\leq0$$ or $$23a^2-14a-1\geq0,$$ which is obvious again for $a\geq1$.