Problem
ALG-B2-M05-P016 Tangent to the square root
#16
★★★★☆ Level 4 of 5
Prove for \(x\ge0\): \[\sqrt{x}\le\frac{x+1}{2}.\]
Hint. Use the tangent to \(\sqrt{x}\) at \(1\), or square both sides.
Since \(\sqrt{x}\) is concave, its graph lies below the tangent at \(1\). This tangent is \(y=1+\frac12(x-1)=\frac{x+1}{2}\). Hence \(\sqrt{x}\le\frac{x+1}{2}\).
A useful linear estimate for problems where roots must be removed.