Toán 12 Tìm giá trị nhỏ nhất

H

huytrandinh

ta có [TEX]\frac{a^{3}}{b+1}+\frac{b+1}{4}+\frac{1}{2}\geq \frac{3}{2}a; \frac{b^{3}}{a+1}+\frac{a+1}{4}+\frac{1}{2}\geq \frac{3}{2}b =>P\geq \frac{5}{4}(a+b)-\frac{3}{2}\geq \frac{5}{2}\sqrt{ab}-\frac{3}{2}=1[/TEX]
vậy minP=1 <=>a=b=1
 
You must log in or register to reply here.
Top Bottom