$VT=5+(\frac ab+\frac ba)+(\frac ac +\frac ca)+(\frac ad+\frac da)+(\frac ae+\frac ea)+(\frac bc+\frac cb)+(\frac bd+\frac db)+(\frac be+\frac eb)+(\frac cd+\frac dc)+(\frac ce+\frac ec)+(\frac de+\frac ed)$
Giả sử $0<p\le a\le b\le c\le d\le e\le q$
Ta có $(\frac ae-\frac pq)(\frac ae-\frac qp)\le0\Leftrightarrow \frac ae+\frac ea\le \frac pq+\frac qp$
Tương tự $\frac bd+\frac db\le \frac pq+\frac qp$
Lại có $(1-\frac ab)(1-\frac be)+(1-\frac ba)(1-\frac eb)\ge0\Leftrightarrow \frac ab+\frac be+\frac ba+\frac eb\le2+\frac ae+\frac ea$
$(1-\frac ad)(1-\frac de)+(1-\frac da)(1-\frac ed)\ge0\Leftrightarrow \frac ad+\frac de+\frac da+\frac ed\le2+\frac ae+\frac ea$
$(1-\frac ac)(1-\frac ce)+(1-\frac ca)(1-\frac ec)\ge0\Leftrightarrow \frac ac+\frac ce+\frac ca+\frac ec\le2+\frac ae+\frac ea$
$(1-\frac bc)(1-\frac cd)+(1-\frac cb)(1-\frac dc)\ge0\Leftrightarrow \frac bc+\frac cd+\frac cb+\frac dc\le2+\frac bd+\frac db$
$\Rightarrow VT\le13+4(\frac ae+\frac ea)+2(\frac bd+\frac db)\le13+6(\frac pq+\frac qp)=VP$
Dấu bằng xảy ra khi $(a;b;c;d;e)=(p;p;p;q;q)\text{ hoặc }(p;p;q;q;q)$ và các hoán vị