Show that 1+xloge⁡x+x2+1≥1+x2 for all x≥0

Let f(x)=1+xloge⁡x+x2+1−1+x2⇒ f′(x)=ln⁡x+x2+1+xx+x2+1 1+2x2x2+1−2x21+x2⇒ f′(x)=1nx+x2+1≥0∀x≥0so, f(x) increases when x≥0hence x≥0⇒f(x)≥f(0) Thus, 1+xln⁡x+x2+1−1+x2≥0 for x≥0  So, 1+xln⁡x+x2+1≥1+x2 for x≥0