Inequality 7320

Let a, b, and c be positive real numbers whose sum is 3, each of which is at most 2.

Prove that the inequality holds:

a2 + b2 + c2 + 3abc <9

Correct answer:

d =  1

Step-by-step explanation:

$$\begin{gathered} a+b+c=3 \\ 0<a<2 \\ 0<b<2 \\ 0<c<2 \\ (a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca \\ (a+b+c{)}^2=3^2=9 \\ {a}^2+b^2+c^2=9-2(ab+bc+ac) \\ 9-2(ab+bc+ac)+3abc<9 \\ -2(ab+bc+ac)<-3abc \\ 2(ab+bc+ac)>3abc \\ a=b=c=2 \\ {x}_{11}=2\cdot (2\cdot 2+2\cdot 2+2\cdot 2)=24 \\ {x}_{12}=3\cdot 2\cdot 2\cdot 2=24 \\ {x}_{11}=x_{12} \\ a=b=c=3/2 \\ {x}_{21}=2\cdot (1.5\cdot 1.5+1.5\cdot 1.5+1.5\cdot 1.5)=13.5 \\ {x}_{22}=3\cdot 1.5\cdot 1.5\cdot 1.5=10.125 \\ {x}_{21}>x_{22} \\ d=1 \end{gathered}$$

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