Question

Buna!
Lectia : Inegalitatea mediilor
1. Fie a,b numere reale pozitive. Sa se arate ca :
a) (a+b)/2 ≤ radical din [(a² + b²)/2]
b) Ma - Mg ≥ Mg - Mh
Ma= media aritmetica
Mg= Media geometrica
Mh= Media armonica
c) Mg ≤ (Ma+ Mh)/2 ≤ Ma


2. Daca 0 a) Ma- Mh ≤ [(b-a)²] / 4a
b) Ma- Mg ≤ [(b-a)²] / 8a

Answer 1

1.a)[tex] \frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}\\ \frac{(a+b)^2}{4}\leq \frac{a^2+b^2}{2}\\ (a+b)^2\leq 2a^2+2b^2\\ a^2+2ab+b^2 \leq 2a^2+2b^2\\ a^2-2ab+b^2\leq 0\\ (a-b)^2\leq 0(A)[/tex]
b)[tex]\frac{a+b}{2}-\sqrt{a*b} \geq \sqrt{a*b}-\frac{2ab}{a+b}\\ \frac{t}{2}- \sqrt{u}\geq\sqrt{u}-\frac{2u}{t}\\ \frac{t}{2}+\frac{2u}{t}\geq 2\sqrt{u}\\ \frac{t^2+4u}{2t}\geq2\sqrt{u}\\ t^2+4u\geq4t\sqrt{u}\\ ( t^2+4u)^2\geq 16t^2u\\ t^4+8t^2u+16u^2\geq16t^2u\\ t^4-8t^2u+16u^2\geq 0\\ (t^2-4u)^2\geq 0(A)[/tex]
2.a)[tex]\frac{a+b}{2}-\frac{2ab}{a+b}\leq \frac{(b-a)^2}{4a} \\ \frac{a^2+2ab+b^2-4ab}{2(a+b)}\leq \frac{(b-a)^2}{4a}\\ \frac{a^2-2ab+b^2}{2(a+b)}\leq \frac{(b-a)^2}{4a}\\
\frac{(a-b)^2}{2(a+b)}\leq \frac{[-(a-b)]^2}{4a}\\ \frac{t^2}{a+b}\leq \frac{t^2}{2a}\\ t^22a\geq t^2(a+b)\\ 2a\geq a+b\\ a\geq b(A) [/tex]
b)Asta nu stiu cum se face...