Câu 1) `bc=a^2`
Ta có: `S=1/ 2 ah_a=1/ 2 b.h_b=1/ 2c.h_c`
`=>h_a={bh_b}/a={ch_c}/a`
`=>h_a^2={bh_b.ch_c}/{a^2}`
`={bch_b.h_c}/{a^2}={a^2.h_b.h_c}/{a^2}=h_b.h_c`
Vậy `h_a^2=h_b.h_c` (đpcm)
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Câu 2.
Vì `a;b;c>0` áp dụng BĐT Cosi ta có:
`1/a +1/ b+1/ c\ge ` $3\sqrt[3]{\dfrac{1}{abc}}$
$a+b+c\ge 3\sqrt[3]{abc}$
`=>(1/a+1/b+1/c)(a+b+c)\ge 3`$\sqrt[3]{\dfrac{1}{abc}} . 3\sqrt[3]{abc}=9$
`=>1/a+1/b+1/c\ge 9/{a+b+c}`
Ta có: $S=pr$ với `p={a+b+c}/2; r` là bán kính đường tròn nội tiếp tam giác$∆ABC$
`=>2S=(a+b+c).r`
Ta lại có:
`\qquad S=1/ 2 ah_a=1/ 2 b.h_b=1/ 2c.h_c`
`=>h_a={2S}/a; h_b={2S}/b; h_c={2S}/c`
`h_a+h_b+h_c=2S (1/ a+1/b+1/c)\ge 2S. 9/{a+b+c}`
`=>h_a+h_b+h_c\ge (a+b+c).r. 9/{a+b+c}=9r`
Dấu "=" xảy ra, tức là `h_a+h_b+h_c=9r` khi $a=b=c$ hay $∆ABC$ đều.
Vậy $h_a+h_b+h_c=9r$ thì $∆ABC$ đều (đpcm)
