Với c
≥0,d
≤0" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;"
>c≥0,d≤0c≥0,d≤0 hoặc c
≤0,d
≥0" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;"
>c≤0,d≥0c≤0,d≥0. Ta có cd
≤0(1)" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;"
>cd≤0(1)cd≤0(1)Với c
≤0,d
≤0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>c≤0,d≤0c≤0,d≤0. Ta có c+d
≤0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>c+d≤0c+d≤0 ( ko thể xảy ra )Với c
≥0,d
≥0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>c≥0,d≥0c≥0,d≥0. Áp dụng bđt AM
−GM" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>AM−GMAM−GM : cd
≤(c+d)24=94(2)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>cd≤(c+d)24=94(2)cd≤(c+d)24=94(2)Từ (1),(2)
⇒cd
≤94" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>(1),(2)⇒cd≤94(1),(2)⇒cd≤94
∀cd" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>∀cd∀cd
∈R" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>∈R∈R~~~~~~~~~~~~~~Áp dụng bđt bunhia, ta có (ac+bd)2
≤(a2+b2)(c2+d2)
⇔|ac+bd|
≤c2+d2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>(ac+bd)2≤(a2+b2)(c2+d2)⇔|ac+bd|≤c2+d2−−−−−−√(ac+bd)2≤(a2+b2)(c2+d2)⇔|ac+bd|≤c2+d2Ta lại có ac+bd
≤|ac+bd|" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>ac+bd≤|ac+bd|ac+bd≤|ac+bd|
⇒ac+bd
≤c2+d2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>⇒ac+bd≤c2+d2−−−−−−√⇒ac+bd≤c2+d2
⇔VT
≤c2+d2+cd=9
−2cd+cd=322.9
−2cd322+cd" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>⇔VT≤c2+d2−−−−−−√+cd=9−2cd−−−−−−√+cd=32√2−−−−√.9−2cd−−−−−−√32√2−−−√+cd⇔VT≤c2+d2+cd=9−2cd+cd=322.9−2cd322+cd
≤322+9
−2cd3222+cd=324+9
−2cd32+cd" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>≤32√2+9−2cd32√22+cd=32√4+9−2cd32√+cd≤322+9−2cd3222+cd=324+9−2cd32+cd=924+32
−232.cd
≤924+32
−232.94" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>=92√4+32√−232√.cd≤92√4+32√−232√.94=924+32−232.cd≤924+32−232.94VT
≤9+624" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>VT≤9+62√4VT≤9+624 (đpcm)Dấu
"=
"" role="presentation" style="font-size: 13.696px; display: inline; position: relative;"
>"=""=" xảy ra
⇔a=b=22,c=d=32" role="presentation" style="font-size: 13.696px; position: relative;"
>⇔a=b=2√2,c=d=32