Ta có:\(\hept{\begin{cases}\\\end{cases}}\)
a + b + c = 7 => b + c = 7 - a
=> 15 = ab + bc + ac = a( b + c ) + bc \(\le a\left(7-a\right)+\frac{\left(b+c\right)^2}{4}\)
<=> \(60\le28a-4a^2+\left(7-a\right)^2\)
<=> \(3a^2-14a+11\le0\)
<=> \(1\le a\le\frac{11}{3}\)
Vậy \(a\le\frac{11}{3}\)
Dấu "=" xảy ra <=> b = c = 5/3
Ta có : \(\hept{\begin{cases}a+b+c=7\\ab+bc+ca=15\end{cases}\Leftrightarrow\hept{\begin{cases}b+c=7-a\\a.\left(b+c\right)+bc=15\end{cases}\Leftrightarrow}\hept{\begin{cases}b+c=7-a\\4.a.\left(b+c\right)+4.b.c=60\end{cases}\left(1\right)}}\)
Với hai số thực b,c ta luôn có : \(\left(b+c\right)^2-4.b.c=\left(b-c\right)^2\ge0\Rightarrow\left(b+c\right)^2\ge4.b.c\Leftrightarrow4.b.c\le\left(b+c\right)^2\left(2\right)\)
Từ ( 1 ) và ( 2) ,ta được : \(60=4.a.\left(b+c\right)+4.b.c\le4.a.\left(7-a\right)+\left(b+c\right)^2=4.a.\left(7-a\right)+\left(7-a\right)^2\)
\(\Leftrightarrow3.a^2-14.a+11\le0\left(a-1\right).\left(3.a-11\right)\le0\)
\(\Leftrightarrow1\le a\le\frac{11}{3}\)(đpcm)
