Question

a⁴+b⁴+c⁴+d⁴>=4abcd

Answer 1

ta có: \(4abcd=2abcd+2abcd\)

ta có:\(\left(ab-cd\right)^2+\left(ac-bd\right)^2\ge0\\ \Leftrightarrow\left(ab\right)^2-2abcd+\left(cd\right)^2+\left(ac\right)^2-2abcd+\left(bd\right)^2\ge0\\ \Leftrightarrow\left(ab\right)^2+\left(cd\right)^2+\left(ac\right)^2+\left(bd\right)^2\ge4abcd\left(1\right)\)

ta có:

\(\left(a^2-b^2\right)^2+\left(a^2-c^2\right)^2+\left(c^2-d^2\right)^2+\left(b^2-d^2\right)^2\ge0\\ \Leftrightarrow a^4-2\left(ab\right)^2+b^4+a^4-2\left(ac\right)^2+c^4+c^2-2\left(cd\right)^2+d^2+b^4-2\left(bd\right)^2+d^4\ge0\\ \Leftrightarrow2\left(a^4+b^4+c^4+d^4\right)\ge2\left(ab+ac+cd+bd\right)\\ \Leftrightarrow a^4+b^4+c^4+d^4\ge ab+ac+cd+bd\left(2\right)\)

từ (1) và (2) \(\Rightarrow a^4+b^4+c^4+d^4\ge4abcd\)

Answer 2

Trần Thị Ngọc Trâm cái (2) với cái (1) chẳng thấy liên quan gì với nhau cả

Answer 3

a mình nhầm chỗ đó\(2\left(a^4+b^4+c^4+d^4\right)\ge2\left(\left(ab\right)^2+\left(ac\right)^2+\left(cd\right)^2+\left(bd\right)^2\right)\\ \Leftrightarrow a^4+b^4+c^4+d^4\ge\left(ab\right)^2+\left(ac\right)^2+\left(cd\right)^2+\left(bd\right)^2\)

cảm ơn ngonhuminh nhé, mình sẽ cẩn thận

Answer 4

ta có: \(a^4+b^4\ge2a^2b^2\\ c^4+d^4\ge2c^2d^2\\ \Rightarrow a^4+b^4+c^4+d^4\ge2\left(ab\right)^2+2\left(cd\right)^2­\left(1\right)\)

ta lại có: \(\left(ab\right)^2+\left(cd\right)^2\ge2abcd\\ \Rightarrow2\left(ab\right)^2+2\left(cd\right)^2\ge4abcd­\left(2\right)\)

từ (1) và (2), suy ra \(a^4+b^4+c^4+d^4\ge4abcd\) (đpcm)