Áp dụng bđt Bunhiacopxki ta có :

\(\left(1+1+1\right)\left[\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\right]\ge\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)^2\)

\(\Leftrightarrow\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\)

\(=\frac{\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\)

Ta lại có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)(bđt quen thuộc; tự cm)

Nên \(\frac{\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{\left(1+\frac{9}{a+b+c}\right)^2}{3}=\frac{10^2}{3}=\frac{100}{3}>\frac{99}{3}=33\)

Hay \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2>33\)(đpcm)

cái này mình nhằm. thay vì bằng 1 nó bằng 11/13 nha. giúp mình với cảm ơn nhiều

\(Q=\frac{a^2+b^2}{a-b}=\frac{\left(a^2-2ab+b^2\right)+2}{a-b}=\frac{\left(a-b\right)^2+2}{a-b}=a-b+\frac{2}{a-b}\)

\(\ge2\sqrt{\left(a-b\right).\frac{2}{a-b}}=2\sqrt{2}\)

mình chẳng hiểu gì cả X_X

Chả hiểu đây là dạng toán gì

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bây h người ta ik hc ik lm , còn bây h bn mới chúc buổi trưa hở . 7:39 ròi nà 

ok Nhung ban phai gui tin nhan cho minh co

Cho vào nước thì CaO tan thành Ca(OH)₂ còn lại CuO

nó chia làm 2 lớp, 1 lớp tan, 1 lớp ko tan(thể rắn) CuO cũng rắn, không lọc được bằng cách này