Problem 1. Let $a,b,c>0$ and $ab+bc+ca+abc=4$
Prove that:
$$\sqrt{ab}+\sqrt{bc}+\sqrt{ca} \leq 3$$
Problem 2. Let $a,b,c \in R$ and $abc=1$
Prove that:
$$\large \frac{1}{(a+1)^2+b^2+1}+\frac{1}{(b+1)^2+c^2+1}+\frac{1}{(c+1)^2+a^2+1} \leq \frac{1}{2}$$
Problem 3. Let $a,b,c>0$
Prove that:
$$\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)} \leq \frac{3}{2}\sqrt{(a+1)(b+1)(c+1)}$$
Prove that:
$$\sqrt{ab}+\sqrt{bc}+\sqrt{ca} \leq 3$$
Problem 2. Let $a,b,c \in R$ and $abc=1$
Prove that:
$$\large \frac{1}{(a+1)^2+b^2+1}+\frac{1}{(b+1)^2+c^2+1}+\frac{1}{(c+1)^2+a^2+1} \leq \frac{1}{2}$$
Problem 3. Let $a,b,c>0$
Prove that:
$$\sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)} \leq \frac{3}{2}\sqrt{(a+1)(b+1)(c+1)}$$
-Sưu tầm-
Để tao làm thử Dũng nha!
ReplyDeleteTao làm bài 2 trước nha:
ReplyDeleteTa có: $(a+b)^2+b^2+1 =a^2+b^2+2a+2\geq 2ab+2a+2$
Tương tự ta có:
$$P \leq \frac{1}{2}[\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}]=\frac{1}{2}$$
với $abc=1$
Dấu "=" xảy ra khi: $a=b=c=1$
Biết tao là ai không dũng. Đứa ngồi cạnh m đây :))).
Dũng đẹp trai ơi. Cho t mần ad với đi :)))
ReplyDelete