Question:
If \(a>0\) then prove that
\((a^3+a^2+a+1)^2≥16a^3\).
Solution:
Let the numbers be, \(a^3, a^2, a, 1\)
We know that A.M. ≥ G.M.
So,
\(\dfrac{a^3+a^2+a+1}{4}≥(a^3 \cdot a^2 \cdot a \cdot 1)^{1/4}\)
\(\implies a^3+a^2+a+1 ≥ 4a^{3/2}\)
Squaring both sides,
\(a^3+a^2+a+1)^2≥16a^3\).
Proved!
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