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Aplicatie-CBS
Posted: Sun Nov 16, 2008 10:43 pm
by Marius Mainea
Fie a,b,c>0. Sa se arate ca:
\( (a^2+b^2)(b^2+c^2)(c^2+a^2)\ge abc(a+b)(b+c)(c+a). \)
Virgil Nicula
Posted: Mon Nov 17, 2008 10:40 pm
by Claudiu Mindrila
Superba inegalitate!
Re: Aplicatie-CBS
Posted: Tue Nov 18, 2008 12:06 pm
by Virgil Nicula
\( \{a,b,c\}\subset\left(0,\infty\right)\ \Longrightarrow\ \left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\ge abc(a+b)(b+c)(c+a) \) .
Demonstratie (la nuvel de clasa a VII - a !) .
\( \left\|\ \begin{array}{c}
a^4+b^2c^2\ge 2a^2bc\\\\
a^2\left(b^2+c^2\right)=a^2\left(b^2+c^2\right)\end{array}\ \right\|\ \bigoplus\ \Longrightarrow\ \left(a^2+b^2\right)\left(a^2+c^2\right)\ge a^2(b+c)^2 . \)
\( \left\|\ \begin{array}{c}
\left(a^2+b^2\right)\left(a^2+c^2\right)\ge a^2(b+c)^2\\\\
\left(b^2+c^2\right)\left(b^2+a^2\right)\ge b^2(c+a)^2\\\\
\left(c^2+a^2\right)\left(c^2+b^2\right)\ge c^2(a+b)^2\end{array}\ \right\|\ \bigodot\ \Longrightarrow\ \prod\left(b^2+c^2\right)\ge abc\prod (b+c) \) .
Observatie. Vezi si aici , unde intr-adevar avem nevoie de inegalitatea C.B.S. !
Posted: Tue Nov 18, 2008 8:00 pm
by Laurian Filip
\( 2(a^2+b^2) \geq (a+b)^2 \)
\( 2(b^2+c^2) \geq (b+c)^2 \)
\( 2(c^2+a^2) \geq (c+a)^2 \)
\( a^2+b^2 \geq 2ab \)
\( b^2+c^2 \geq 2bc \)
\( c^2+a^2 \geq 2ca \)
prim inmultire
\( 8 ((a^2+b^2)(b^2+c^2)(c^2+a^2))^2 \geq 8 (abc(a+b)(b+c)(c+a))^2 \)
echivalent cu
\( (a^2+b^2)(b^2+c^2)(c^2+a^2) \geq abc(a+b)(b+c)(c+a) \)