Fie \( a,b,c \) trei numere nenegative . Sa se demonstreze ca :
\( (a^2-bc)\sqrt{b+c}+(b^2-ca)\sqrt{c+a}+(c^2-ab)\sqrt{a+b}\ge 0 \)
Inegalitate in numere nenegative
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Inegalitate in numere nenegative
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Mateescu Constantin
- Newton
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Notam cu \( A=\sqrt{b+c} \) si analoagele.
Astfel, inegalitatea se scrie:
\( A(a^2-bc)+B(b^2-ca)+C(c^2-ab)\ge 0 \)
Avem \( 2\sum A(a^2-bc)=\sum A[(a-b)(a+c)+(a-c)(a+b)]=\sum A(a-b)(a+c)+\sum B(b-a)(b+c)=\sum (a-b)[A(a+c)-B(b+c)]=\sum(a-b)\frac{A^2(a+c)^2-B^2(b+c)^2}{A(a+c)+B(b+c)}=\sum \frac{(a-b)^2(a+c)(b+c)}{A(a+c)+B(b+c)}\ge 0 \).
Astfel, inegalitatea se scrie:
\( A(a^2-bc)+B(b^2-ca)+C(c^2-ab)\ge 0 \)
Avem \( 2\sum A(a^2-bc)=\sum A[(a-b)(a+c)+(a-c)(a+b)]=\sum A(a-b)(a+c)+\sum B(b-a)(b+c)=\sum (a-b)[A(a+c)-B(b+c)]=\sum(a-b)\frac{A^2(a+c)^2-B^2(b+c)^2}{A(a+c)+B(b+c)}=\sum \frac{(a-b)^2(a+c)(b+c)}{A(a+c)+B(b+c)}\ge 0 \).