Нужно решение!! С объяснениями!!

0\\\\ \frac{9t}{\sqrt{t^2+5}}-\frac{\sqrt{t^2+5}}{2} \geq \sqrt{6t}\\\\ 18t-(t^2+5) \geq 2\sqrt{6t^3+30t}\\\\ -t^2+18t-5 \geq 2\sqrt{6t^3+30t} \\\\ (-t^2+18t-5)^2 -4(6t^3+30t) \geq 0 \\\\\ " alt="9(1+5^{1-2x})^{-\frac{1}{2}}-\frac{1}{2}(5^{2x}+5)^{\frac{1}{2}} \geq 6^{\frac{1}{2}}*5^{\frac{x}{2}}\\\\ \frac{9}{\sqrt{1+\frac{5}{5^{2x}}}} - \frac{\sqrt{5^{2x}+5}}{2} \geq \sqrt{ 6*5^x}\\\\ 5^{1-2x} \neq -1\\\\ x\in(-\infty; + \infty)\\\\ 5^x=t>0\\\\ \frac{9t}{\sqrt{t^2+5}}-\frac{\sqrt{t^2+5}}{2} \geq \sqrt{6t}\\\\ 18t-(t^2+5) \geq 2\sqrt{6t^3+30t}\\\\ -t^2+18t-5 \geq 2\sqrt{6t^3+30t} \\\\ (-t^2+18t-5)^2 -4(6t^3+30t) \geq 0 \\\\\ " align="absmiddle" class="latex-formula">
$(t-1)(t-5)(t^2-54t+5) \geq 0\\ 5^x=1\\ 5^x=5\\ x=0\\ x=1\\\\ x\in[0;1]$