Question

[Edogawa Conan]

1. CM: \(3\left(a^2+b^2\right)-ab+4\ge2\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\)

2. CMR: \(a^4+b^4+c^4+1\ge2a\left(ab^2-a+c+1\right)\)

3. Cm: \(\left(a^5+b^5\right)\left(a+b\right)\ge\left(a^4+b^4\right)\left(a+b\right)\)

Answer 1

[Đặng Ngọc Quỳnh]

1. BĐT tương đương với \(6\left(a^2+b^2\right)-2ab+8-4\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\ge0\)

\(\Leftrightarrow\left[a^2-4a\sqrt{b^2+1}+4\left(b^2+1\right)\right]+\left[b^2-4b\sqrt{a^2+1}+4\left(a^2+1\right)\right]\)\(+\left(a^2-2ab+b^2\right)\ge0\)

\(\Leftrightarrow\left(a-2\sqrt{b^2+1}\right)^2+\left(b-2\sqrt{a^2+1}\right)^2+\left(a-b\right)^2\ge0\)(đúng)

=> Đẳng thức không xảy ra

Answer 2

[Đặng Ngọc Quỳnh]

2. \(a^4+b^4+c^2+1\ge2a\left(ab^2-a+c+1\right)\)

\(\Leftrightarrow a^4+b^4+c^2+1\ge2a^2b^2-2a^2+2ac+2a\)

\(\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+\left(c^2-2ac+a^2\right)+\left(a^2-2a+1\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c-a\right)^2+\left(a-1\right)^2\ge0\)