Prove that x^6+x^4+x^2+x+3=0 has no positive real roots.?

Applying Descartes' rule of signs:
When x takes positive value, the algebraic signs of all terms of the function are positive; and there is no change in sign. So it has no positive real roots.

When x takes negative value, the signs of the terms are: +, +, +, -, +. So there are two changes in sign; hence it has at most two negative real roots.

Rest of the roots are all imaginary, which will be in even numbers only.
Thus the system has - either 2 negative real roots and 4 complex roots
(or) all the 6 complex roots.

No positive real roots.

For any real number x :
x² ≥ 0
x⁴ = (x²)² ≥ 0
x⁶ = (x³)² ≥ 0

Hence, x⁶ + x⁴ + x² ≥ 0
and -(x⁶ + x⁴ + x²) ≤ 0

x⁶ + x⁴ + x² + x + 3 = 0
x = -(x⁶ + x⁴ + x²) - 3

Since -(x⁶ + x⁴ + x²) ≤ 0
x = -(x⁶ + x⁴ + x²) - 3 ≤ -3

Hence, x has no positive real roots.

The left-hand side is positive for any non-negative input.