Chứng minh

A

anhnhduc001

V

viethoang1999

2)

Dùng Cauchy ngược :
$\dfrac{a}{1+b^{2}c}=a-\dfrac{ab^{2}c}{1+b^{2}c}$ \geq $a-\dfrac{ab^{2}c}{2b\sqrt{c}}=a-\dfrac{ab\sqrt{c}}{2}=a-\dfrac{b\sqrt{ac.a}}{2}$ \geq $a-\dfrac{b(ac+a)}{4}=a-\dfrac{ab+abc}{4}$
Thiết lập các BĐT tương tự rồi cộng vế :
$VT$ \geq $(a+b+c+d)-\dfrac{ab+bc+cd+da+abc+bcd+cda+dab}{4}$
Mặt khác, áp dụng BĐT AM-GM :
$ab+bc+cd+da=(a+c)(b+d)$ \leq $\dfrac{1}{4}(a+b+c+d)^{2}=4$
$abc+bcd+cda+dab=bc(a+d)+da(b+c)$ \leq $\dfrac{(b+c)^{2}(a+d)}{4}+\dfrac{(d+a)^{2}(b+c)}{4}=\dfrac{1}{4}(b+c)(a+d)(a+b+c+d)$ \leq $\dfrac{1}{16}(a+b+c+d)^{3}=4$
Suy ra $VT$ \geq $4-\dfrac{4+4}{4}=2$ $(đ.p.c.m)$
Đẳng thức xảy ra khi $a=b=c=1$



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V

viethoang1999

1)

$\sum \dfrac{a+1}{b^2+1}=\sum (a+1)-\sum \dfrac{(a+1)b^2}{b^2+1}$ \geq $\sum (a+1)-\sum \dfrac{(a+1)b^2}{2b}=\sum (a+1)-\sum \dfrac{(a+1)b}{2}=3+\dfrac{\sum a-\sum ab}{2}$ \geq $3$

Dấu "=" xảy ra khi: $a=b=c=1$


“Bài dự thi event box toán 10”
 
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