Question

Se consideră numerele reale a şi b.Dacă $a^{2} + b^{2} -a-5b+5=0$ , sa se arate ca a+b ∈ [ 1;5]

Answer 1

$Inmultim~relatia~cu~4: \\ \\ 4a^2+4b^2-4a-20b+20=0.~Adunand~6~in~ambii~membri,~si~ \\ \\ grupand~convenabil,~avem:~(4a^2-4a+1)+(4b^2-20b+25)=6 \Leftrightarrow \\ \\ \Leftrightarrow (2a-1)^2+(2b-5)^2=6. \\ \\ Folosim~inegalitatea~2(x^2+y^2) \geq (x+y)^2,~adevarata~ \forall~x,y \in R,~si \\ \\ obtinem: ~2 \Big((2a-1)^2+(2b-5)^2 \Big) \geq (2a-1+2b-5)^2 \Leftrightarrow 12 \geq \\ \\ \geq (2a+2b-6)^2 .$

$Rezulta:~ - \sqrt{12} \leq 2a+2b-6 \leq \sqrt{12} \Rightarrow a+b \in \Big(3- \sqrt{3}~;~3+ \sqrt{3} \Big). \\ \\ \Big(3- \sqrt{3}~;~3+ \sqrt{3} \Big) \subset [1;5] \Rightarrow a+b \in [1;5].$