Logarithmic Inequality and quadratic function

Question

Prove that if $m$ satisfy inequality: $$\left (1+ \frac{1}{2m}\right) \log_k 3 - \log_k (27 + 3^\frac{1}{2m}) \le 2$$ where $k=\frac{1}{2}$, then $x^2 + mx + 1 > 0$ for every real number. I solved one $m_1=0$, but the other doesn't satisfy the second condition.

Answer 1

HINT.-You have $$\left (1+ \frac{1}{2m}\right) \log_k x - \log_k (27 + 3^\frac{1}{2m}) \le 2\iff\log_k\frac{x^{1+\frac{1}{2m}}}{27+ 3^{\frac{1}{2m}}}\le2=\log_k\frac14$$ and because the log is injective and increasing we get $$\frac{x^{1+\frac{1}{2m}}}{27+ 3^{\frac{1}{2m}}}\lt\frac14$$ Now you can work in an easier way.Try it and if you can't finish I shall give you the final part.