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Demonstrați următoarele inegalitati, in ipotezele indicate:
a^2 + b^2 mai mare sau egal decat 2
Daca a + b =2

Răspuns :

[tex]\text{Aplicam inegalitatea dintre media aritmetica si cea patratica:}\\\sqrt{\dfrac{a^2+b^2}{2}} \geqslant \dfrac{a+b}{2}\\\sqrt{\dfrac{a^2+b^2}{2}} \geqslant \dfrac{2}{2}\\\sqrt {\dfrac{a^2+b^2}{2}} \geqslant 1 | ()^2\\\dfrac{a^2+b^2}{2} \geqslant 1 |\cdot 2\\a^2+b^2 \geqslant 2 ,Q.E.D.[/tex]

Mocanualexandrp2ikb6

Folosim ipoteza a+b=2 <=> a²+b² >=a+b ,aplicam inegalitatea C-B-S si avem

(1·a+1·b)² <=(1²+1²)·(a²+b²) <=> (a+b)² <=2·(a²+b²) <=> a²+b² >=(a+b)²/2 <=>

a²+b² >=4/2=2 <=> a²+b² >=2 .

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