In legatura cu inegalitatea C.B.S.

[Claudiu Mindrila]

\( 1. \)
Demonstrati ca \( \forall a,b \in \mathbb{R} \), are loc inegalitatea:
\( (a^2+1)(b^2+1)\geq(a+b)(ab+1) \).
Valentin Vornicu, concursul "Nicolae Paun", 2003

\( 2. \)
Demonstrati ca daca \( a,b,c,d>0 \) astfel incat \( a^2+b^2+c^2+d^2=4. \), atunci \( a^3+b^3+c^3+d^3\geq 4. \)

[Laurian Filip]

1. din cbs \( (a^2+1)(b^2+1) \geq (ab+1)^2 \)
\( (ab-1)^2 \geq 0 \)
\( (ab-1)^2+(a+b)^2 \geq (a+b)^2 \)
\( (a^2+1)(b^2+1) \geq (a+b)^2 \)

inmultind cele 2 relatii obtinem
\( (a^2+1)(b^2+1)\geq(a+b)(ab+1) \)



2. \( (a-1)^2+(b-1)^2+(c-1)^2+(d-1)^2\geq 0 \)
\( a^2+b^2+c^2+d^2 +4 \geq 2(a+b+c+d) \)
\( 2(a^2+b^2+c^2+d^2) \geq 2(a+b+c+d) \)
\( (a^2+b^2+c^2+d^2) \geq (a+b+c+d) \)

din cbs

\( (a^3+b^3+c^3+d^3)(a+b+c+d) \geq (a^2+b^2+c^2+d^2) \)

impreuna cu relatia de mai sus,
\( a^3+b^3+c^3+d^3 \geq a^2+b^2+c^2+d^2 =4 \)

[Claudiu Mindrila]

Asa este Filip .
De fapt, mai general, avem ca:
\( (a_1^2+a_2^2+...+a_n^2)^3\leq n^2(a_1^3+a_2^3+...+a_n^3)^2 \).
Intr-adevar, conform inegalitatii Cauchy-Buniakowski-Schwarz avem:

\( \left(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\right)^{2}=\left(\sqrt{a_{1}}\cdot\sqrt{a_{1}^{3}}+\sqrt{a_{2}}\cdot\sqrt{a_{2}^{3}}+...+\sqrt{a_{n}}\cdot\sqrt{a_{n}^{3}}\right)^{2}\leq\left(a_{1}+a_{2}+...+a_{n}\right)\left(a_{1}^{3}+a_{2}^{3}+...+a_{n}^{3}\right) (1) \).
Apoi
\( \left(a_{1}+a_{2}+...+a_{n}\right)^{2}=\left(a_{1}\cdot1+a_{2}\cdot1+...+a_{n}\cdot1\right)^{2}\leq\left(1+1+...+1\right)^{2}\left(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\right)=n^{2}\left(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\right) (2). \)
Din inmultirea inegalitatilor \( (1) \) si \( (2) \) rezulta inegalitatea dorita.
In cazul nostru, pentru \( n=4 \) si \( a_1^2+a_2^2+a_3^2+a_4^2=4 \) rezulta imediat \( 2. \)

[Virgil Nicula]

Demonstrati ca \( \forall\ \{a,b,c \}\subset \mathbb{R} \), are loc inegalitatea : \(
\left(a^2+c^2\right)\left(b^2+c^2\right)\geq c(a+b)(ab+c^2) \) .

Aplicam de doua ori inegalitatea C.B.S. :

\( \left\|\ \begin{array}{c}
(a\cdot c+c\cdot b)^2\le \left(a^2+c^2\right)\left(c^2+b^2\right)\\\\
(a\cdot b+c\cdot c)^2\le \left(a^2+c^2\right)\left(b^2+c^2\right)\end{array}\ \right\|\ \bigodot\ \Longrightarrow\ \left|c(a+b)\left(ab+c^2\right)\right|\ \le\ \left(a^2+c^2\right)\left(b^2+c^2\right) \) .

Dar \( x\le |x| \) . In concluzie, \( c(a+b)\left(ab+c^2\right)\ \le\ \left(a^2+c^2\right)\left(b^2+c^2\right) \) . Avem egalitate

daca si numai daca \( a=b=c \) . Pentru \( c\ :=\ 1 \) se obtine inegalitatea propusa initial.

[Claudiu Mindrila]

Pentru \( 1. \), din inegalitatea C.B.S. avem:
\( (a^2+1)(b^2+1)\geq (ab+1)^2 \)
si
\( (a^2+1)(1+b^2)\geq (a+b)^2 \).
Din inmultirea acestor inegalitati rezulta cerinta.