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I just started reading the Introduction to Algorithms textbook (CLRS) and they introduced the substitution method when solving Divide and Conquer algorithm. In the final statement, I am confused as in the last step, why does T(n)<=dn^2 holds as long as d >= (16/13)c?. Can someone explain?

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why does $\;T(n) \le d \,n^2\;$ hold as long as $\;d \ge \cfrac{16}{13}\,c\;$?

If $\;c \le \cfrac{13}{16}d\;$ then $\cfrac{3}{16}dn^2+c n^2 \le \cfrac{3}{16}dn^2 + \cfrac{13}{16}d n^2 = \left(\cfrac{3}{16}+\cfrac{13}{16}\right)dn^2 = dn^2$

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    thank you! so simple after u explain it – 2017-01-25