Răspuns:
Explicație pas cu pas:
O posibila demonstratie:
[tex]\texttt{Fie }\log_ab=a,~\log_bc=b,~\log_ca=c\\\texttt{Sa observam ca }a\cdot b\cdot c=1\\\texttt{Inegalitatea devine :}\\\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\leq\dfrac{a^2+b^2+c^2}{2}\\\texttt{Sa observam ca in loc de 1 putem scrie }a\cdot b\cdot c:\\\dfrac{a\cdot b\cdot c}{a+b}+\dfrac{a\cdot b\cdot c}{b+c}+\dfrac{a\cdot b\cdot c}{c+a}\leq\dfrac{a^2+b^2+c^2}{2}\\\texttt{Sa observam ca}:[/tex]
[tex]\dfrac{a+b}{a\cdot b\cdot c}=\dfrac{1}{b\cdot c}+\dfrac{1}{a\cdot c}\stackrel{C.B.S}{\geq}\dfrac{(1+1)^2}{bc+ac}=\dfrac{4}{bc+ac}\\\texttt{Prin urmare }\dfrac{a\cdot b\cdot c}{a+b}\leq \dfrac{bc+ac}{4}\\\texttt{Analog:}\dfrac{a\cdot b\cdot c}{b+c}\leq\dfrac{ab+ac}{4}\\~~~~~~~\dfrac{a\cdot b\cdot c}{a+c}\leq\dfrac{ab+bc}{4}\\\texttt{Adunand inecuatiile obtinem:}\\L.H.S\leq\dfrac{2(ab+bc+ac)}{4}=\dfrac{ab+ac+bc}{2}\\\texttt{Mai avem de demonstrat ca:}\\\dfrac{ab+ac+bc}{2}\leq \dfrac{a^2+b^2+c^2}{2}[/tex]
[tex]\texttt{Prin inmultirea ecuatiei cu 2 obtinem:}\\ab+ac+bc\leq a^2+b^2+c^2\\\texttt{Asta e o inegalitate destul de cunoscuta.}\\\texttt{Mai inmultim odata cu 2:}\\2ab+2bc+2ac\leq 2a^2+2b^2+2c^2\\2a^2+2b^2+2c^2-2ab-2bc-2ac\geq 0\\(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)\geq 0\\(a-b)^2+(b-c)^2+(a-c)^2\geq 0,~~~\texttt{Q.E.D.}[/tex]
