xét$:(a+b+c)(1/a+1/b+1/c)=1+a/b+a/c+b/a+1+b/c+c/a+c/b+1=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)$do $a,b,c$ dương nên$:a/b+b/a\geq2$,tương tự 2 cặp kia cũng thế$\Rightarrow $(a+b+c)(1/a+1/b+1/c)⩾9:(a+b+c)(1/a+1/b+1/c)⩾9" role="presentation" style="font-size: 13.696px; word-spacing: 0px; position: relative;">:(a+b+c)(1/a+1/b+1/c)⩾9
xét$:(a+b+c)(1/a+1/b+1/c)=1+a/b+a/c+b/a+1+b/c+c/a+c/b+1=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)$do $a,b,c$ dương nên$:a/b+b/a\geq2$,tương tự 2 cặp kia cũng thế$\Rightarrow đpcm$
xét$:(a+b+c)(1/a+1/b+1/c)=1+a/b+a/c+b/a+1+b/c+c/a+c/b+1=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)$do $a,b,c$ dương nên$:a/b+b/a\geq2$,tương tự 2 cặp kia cũng thế$\Rightarrow
$(a+b+c)(1/a+1/b+1/c)⩾9:(a+b+c)(1/a+1/b+1/c)⩾9" role="presentation" style="font-size: 13.696px; word-sp
ac
ing: 0px; position: relative;">:(a+b+c)(1/a+1/b+1/c)⩾9