Cho 3 số dương a,b,c thỏa mãn `a+b ≤ c` . CMR : `(a^2 + b^2 + c^2)(1/a^2 + 1/b^2 + 1/c^2) ≥ 27/2`

Với `a;b;c>0` ta có:

`\qquad (a-b)^2\ge 0`

`<=>a^2+b^2-2ab\ge 0`

`<=>a^2+2ab+b^2-4ab\ge 0`

`<=>(a+b)^2\ge 4ab`

$\\$

Vì `0<a+b\le c`

`=>c^2\ge (a+b)^2\ge 4ab` 

Đặt

`A=(a^2+b^2+c^2)(1/{a^2}+1/{b^2}+1/{c^2})`

`=1+{a^2}/{b^2}+{a^2}/{c^2}+{b^2}/{a^2}+1+{b^2}/{c^2}+c^2.(1/{a^2}+1/{b^2})+1`

`=3+({a^2}/{b^2}+{b^2}/{a^2})+({a^2}/{c^2}+{c^2}/{16a^2})+({b^2}/{c^2}+{c^2}/{16b^2})+{15c^2}/{16}. (1/{a^2}+1/{b^2})`

Áp dụng BĐT Cosi ta có:

`{a^2}/{b^2}+{b^2}/{a^2}\ge 2\sqrt{{a^2}/{b^2} .{b^2}/{a^2}}=2`

`{a^2}/{c^2}+{c^2}/{16a^2}\ge 2\sqrt{{a^2}/{c^2} .{c^2}/{16a^2}}=1/ 2`

`{b^2}/{c^2}+{c^2}/{16b^2}\ge 2\sqrt{{b^2}/{c^2} .{c^2}/{16b^2}}=1/ 2`

`1/{a^2}+1/{b^2}\ge 2\sqrt{1/{a^2} . 1/{b^2}}=2/{ab} `

$\\$

`=>A\ge 3+2+1/2+1/2+{15}/{16}. 4ab . 2/{ab}`

`\qquad ` (do `c^2\ge 4ab)`

`=>A\ge 6+{15}/2={27}/2`

Dấu “=” xảy ra khi `a=b=c/2`

Vậy với `a;b;c` dương thỏa `a+b\le c` ta có: `(a^2+b^2+c^2)(1/{a^2}+1/{b^2}+1/{c^2})\ge {27}/2`