Ta cần chứng minh: \(3\left(a^2+b^2\right)+c^2\ge2\left(ab+bc+ca\right)\)

Nó đúng bởi \(3\left(a^2+b^2\right)+c^2-2\left(ab+bc+ca\right)=\left(a-b\right)^2+2\left(a-\frac{c}{2}\right)^2+2\left(b-\frac{c}{2}\right)^2\ge0\)

Đẳng thức xảy ra khi \(a=b=\frac{1}{\sqrt{5}};c=\frac{2}{\sqrt{5}}\)

Done!

ĐK: \(x\ge1\)Bpt \(\Leftrightarrow\sqrt{x^2+2x+92}-10\ge\left(x^2+2x-8\right)+\left(\sqrt{x-1}-1\right)\)

\(\Leftrightarrow\frac{x^2+2x-8}{\sqrt{x^2+2x+92}+10}\ge\left(x-2\right)\left(x+4\right)+\frac{x-2}{\sqrt{x-1}+1}\)

\(\Leftrightarrow\left(x-2\right)\left[\frac{x+4}{\sqrt{x^2+2x+92}+10}-\left(x+4\right)-\frac{1}{\sqrt{x-1}+1}\right]\ge0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x+4\right)\left(\frac{1}{\sqrt{x^2+2x+92}+10}-1\right)-\frac{1}{\sqrt{x-1}+1}\right]\ge0\)

Ta có: \(\left(x+4\right)\left(\frac{1}{\sqrt{x^2+2x+92}+10}-1\right)-\frac{1}{\sqrt{x-1}+1}< 0\left(\forall x\ge1\right)\)

Do đó bpt \(\Leftrightarrow x-2\le0\Leftrightarrow x\le2\)

Kết hợp với ĐK ta có nghiệm của bpt là:\(1\le x\le2\)

ĐK: \(x\ge1\)

\(BPT\Leftrightarrow x^2-2x+3x-6+\sqrt{x-1}-1+x+8-\sqrt{x^2+2x+82}\le0\)

\(\Leftrightarrow\left(x-2\right)\left[x+3+\frac{1}{\sqrt{x-1}+1}+\frac{14}{x+8+\sqrt{x^2+2x+92}}\right]\le0\)

Mà \(x+3+\frac{1}{\sqrt{x-1}+1}+\frac{14}{x+8+\sqrt{x^2+2x+92}}>0\forall x\ge1\)

\(BPT\Leftrightarrow x-2\le0\Leftrightarrow x\le2\)

ĐK \(x\ge-1\)

\(BPT\Leftrightarrow\frac{2}{3}x-\sqrt{x+1}+\frac{4}{3}x+2-2\sqrt{2x+3}\le0\)

\(\Leftrightarrow\left(x-3\right)\left[\frac{2\left(4x+3\right)}{3\left(2x+3\sqrt{x+1}\right)}+\frac{4\left(2x+3\right)}{3\left(2x+3+3\sqrt{2x+3}\right)}\right]\le0\)

\(\Leftrightarrow x-3\le0\Leftrightarrow x\le3\)

= -1090 nha bạn

hello

helo mn ??