Chủ đề này có 1 trả lời

[darksoul]

Cho $a.b.c.d \in [\frac{1}{2};\frac{2}{3}]$. Tìm GTLN:

T=$16(\frac{a+c}{a+d})^2+25(\frac{c+d}{a+b})^2$

[DOTOANNANG]

Ta sẽ chứng minh $\text{T}= \text{f}\,\,\left ( \,\,a,\,\,b,\,\,c,\,\,d\,\, \right )\leqq \text{f}\,\,\left ( \,\,\frac{1}{2},\,\,\frac{1}{2},\,\,\frac{2}{3},\,\,\frac{2}{3}\,\, \right )= \frac{544}{9}$

Xét $d\geqq c\,\,\rightarrow$

$\frac{544}{9}- 16\left ( \frac{a+ c}{a+ d} \right )^{2}- 25\left ( \frac{c+ d}{a+ b} \right )^{2}= \frac{100\left ( 2\,a+ 2\,b+ 3\,d \right )\left [ \left ( a+ b \right )\left ( 2/\,3- d \right )+ d\left ( a- 1/\,2+ b- 1/\,2 \right ) \right ]}{3\left ( a+ b \right )^{2}}$ $+ \frac{\text{Q}\left ( d- c \right )}{\left ( a+ b \right )^{2}\left ( a+ d \right )^{2}}$

 

với: $\text{Q}= 32\,a^{3}+ 64\,a^{2}b+ 41\,a^{2}c+ 91\,a^{2}d+ 32\,ab^{2}+ 32\,abc+ 32\,abd$ $+ 50\,acd+ 150\,ad^{2}+ 16\,b^{2}c+ 16\,b^{2}d+ 25\,cd^{2}+ 75\,d^{3}$

 

Xét $d\leqq c\,\,\rightarrow \,\,16\left ( \frac{a+ c}{a+ d} \right )^{2}+ 25\left ( \frac{c+ d}{a+ b} \right )^{2}\geqq 16 \left ( \frac{a+ c}{a+ c} \right )^{2}+ 25\left ( \frac{1}{\frac{4}{3}} \right )^{2}= \frac{481}{16}$