I tried solving the above question but was unable to prove it. I used Descartes rule of sign, factorisation techniques, and many other things but could not figure out the solution.
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$\begingroup$First note$$2x^6+12x^5+30x^4+60x^3+80x^2+30x+45=2(x^3+3x^2)^2+12\left(x^2+\tfrac52 x\right)^2+5(x+3)^2.$$Not only is this non-negative, but it could only be zero if$$x^3+3x^2=x^2+\tfrac52 x=x+3=0.$$The last condition simplifies to $x=-3$, which contradicts the second condition.
$\endgroup$ 9 $\begingroup$Factor $x^4$ and separate a non-negative part of the expression covering completely the terms $x^6$ and $x^5,$ then factor $x^2,$ ...$$\begin{aligned}P(x)=&2x^6+12x^5+30x^4+60x^3+80x^2+30x+45\\=&2x^4(x^2+6x+9)+12x^4+60x^3+80x^2+30x+45\\=&2x^4(x^2+6x+9)+3x^2(4x^2+20x+25)+30x^2+30x+45\\=&2x^4(x^2+6x+9)+3x^2(4x^2+20x+25)+15(2x^2+2x+3)\end{aligned}$$ which is strictly positive for any real $x.$
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