Câu 1: Số nghiệm là 1 nghiệm

Câu 4: B

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

khi đó:

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

\(=\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\)

Lại có: \(\frac{1}{a}+\frac{1}{b}\ge\frac{\left(1+1\right)^2}{a+b}=\frac{4}{a+b}\)=> \(\frac{2}{a+b}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

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

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

Dấu "=" xảy ra <=> a = b = c = 1

Vậy max P = 3 tại a = b = c =1.

Không thích làm cách này đâu nhưng đường cùng rồi nên thua-_-

Đặt \(\sqrt{x+y}=a;\sqrt{y+z}=b;\sqrt{z+x}=c\) suy ra

\(x=\frac{a^2+c^2-b^2}{2};y=\frac{a^2+b^2-c^2}{2};z=\frac{b^2+c^2-a^2}{2}\). Ta cần chứng minh:

\(abc\left(a+b+c\right)\ge\left(a+b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

Đây là bất đẳng thức Schur bậc 3, ta có đpcm.

         \(a>2b+3\)

\(\Leftrightarrow\)\(4a>8b+12\)

\(\Leftrightarrow\)\(4a-5>8b+12-5\)

\(\Leftrightarrow\)\(4a-5>8b+7\) (đpcm)

        \(a^2+b^2+c^2\ge ab+bc+ca\)

 \(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca\ge0\)

\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

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

Dấu   "="  xảy ra  <=>  a = b = c

+) Bài bất đẳng thức:

\(\dfrac{2017a-a^2}{bc}=\dfrac{\left(a+b+c\right)a-a^2}{bc}=\dfrac{ab+ca}{bc}=\dfrac{a}{c}+\dfrac{a}{b}\left(1\right)\)

Tương tự: \(\left\{{}\begin{matrix}\dfrac{2017b-b^2}{ca}=\dfrac{b}{a}+\dfrac{b}{c}\left(2\right)\\\dfrac{2017c-c^2}{ab}=\dfrac{c}{a}+\dfrac{c}{b}\left(3\right)\end{matrix}\right.\)

\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow\dfrac{2017a-a^2}{bc}+\dfrac{2017b-b^2}{bc}+\dfrac{2017c-c^2}{ab}=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)

\(\sqrt{2}\left(\sum\sqrt{\dfrac{2017-a}{a}}\right)=\sqrt{2}\left(\sum\sqrt{\dfrac{\left(a+b+c\right)-a}{a}}\right)=\sqrt{2}\left(\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}+\sqrt{\dfrac{a+b}{2}}\right)\)

Bất đẳng thức cần chứng minh tương đương với:

\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge\sqrt{2}\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)\)

*Có: \(\sqrt{2.\dfrac{a+b}{c}}+\sqrt{2.\dfrac{b+c}{a}}+\sqrt{2.\dfrac{c+a}{b}}\le\dfrac{2+\dfrac{a+b}{c}}{2}+\dfrac{2+\dfrac{b+c}{a}}{2}+\dfrac{2+\dfrac{c+a}{b}}{2}=3+\dfrac{\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}}{2}\)

Ta chỉ cần chứng minh:

\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge3+\dfrac{\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}}{2}\)

hay \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\) (cái này chị tự chứng minh nhé)

Anh Trần Tuấn Hoàng giỏi BĐT quá nhỉ