\(\frac{1}{\sqrt{1+a^2}}=\frac{\sqrt{bc}}{\sqrt{bc+a.abc}}=\frac{\sqrt{bc}}{\sqrt{bc+a\left(a+b+c\right)}}=\frac{\sqrt{bc}}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)

Tương tự và cộng lại \(\Rightarrow P\le\frac{3}{2}\)

Đẳng thức xảy ra khi \(a=b=c=\sqrt{3}\)

chưa tìm thấy bài nào thực sự khó ơ câu hỏi hay cả buồn thật 

Khó 😩 hay suy nghỉ mà đau 🦁🦁🦁🦁

Cần chứng minh: \(\sqrt{a^2-ab+b^2}\ge\frac{1}{2}\left(a+b\right)\)

Thật vậy: \(\sqrt{a^2-ab+b^2}\ge\frac{1}{2}\left(a+b\right)^2\Leftrightarrow4\left(a^2-ab+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow4a^2-4ab+4b^2-a^2-b^2-2ab\ge0\Leftrightarrow3\left(a^2+b^2-2ab\right)\ge0\Leftrightarrow3\left(a-b\right)^2\ge0\)(đúng)

Áp dụng:\(P=\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ac+a^2}}\)

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

Dấu "=" xảy ra khi: \(a=b=c=1\)

Ta có:

\(P=\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ca+a^2}}\)

\(=\frac{1}{\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}}+\frac{1}{\sqrt{\frac{1}{4}\left(b+c\right)^2+\frac{3}{4}\left(b-c\right)^2}}+\frac{1}{\sqrt{\frac{1}{4}\left(c+a\right)^2+\frac{3}{4}\left(c-a\right)^2}}\)

\(\le2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(\le2.\frac{1}{4}.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)

\(=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)

Bài 1 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)

\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)

b) Để \(A< -1\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)

\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)

\(\Leftrightarrow2\sqrt{x}< 1\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)

\(\Leftrightarrow x< \frac{1}{4}\)

Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)

\(a^2-ab+b^2=\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2\ge\frac{1}{4}\left(a+b\right)^2\)

\(\frac{1}{\sqrt{a^2-ab+b^2}}\le\frac{2}{a+b}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)