\(\left\{{}\begin{matrix}x+y+z=0\\xy+yz+zx=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\2\left(xy+yz+zx\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2xy+2yz+2zx=0\\2xy+2yz+2zx=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx-2xy-2yz-2zx=0\)

\(\Rightarrow x^2+y^2+z^2=0\)

\(\left\{{}\begin{matrix}x^2\ge0\forall x\\y^2\ge0\forall y\\z^2\ge0\forall z\end{matrix}\right.\)

Nên: \(x^2+y^2+z^2\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x^2=0\\y^2=0\\z^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)

Vậy \(x=y=z=0\)

Ta có điều phải chứng minh

\(a=x^2-2x+15=x^2-2x+1+14=\left(x-1\right)^2+14>0\)

\(b=4a^2-4a+23=4a^2-4a+4+19=\left(2a-2\right)^2+19>0\)

\(c=2x^2-7x+37=2\left(x^2-\dfrac{7}{2}x+\dfrac{37}{2}\right)\)

\(c=2\left(x^2-\dfrac{7}{2}x+\dfrac{49}{16}+\dfrac{247}{16}\right)\)

\(c=2\left(x^2-\dfrac{7}{2}x+\dfrac{49}{16}\right)+\dfrac{247}{8}\)

\(c=2\left(x-\dfrac{7}{4}\right)^2+\dfrac{247}{8}>0\)

\(d=4x-9x^2-17=-9x^2+4x-17\)

\(d=-9x^2+4x-\dfrac{4}{9}-\dfrac{149}{9}\)

\(d=-\left(9x^2-4x+\dfrac{4}{9}\right)-\dfrac{149}{9}\)

\(d=-\left(3x-\dfrac{2}{3}\right)^2-\dfrac{149}{9}< 0\)

\(A=\left(x-2\right)\left(x+2\right)=x^2-4\ge-4\)

Dấu"=" xảy ra khi:

\(x=0\)

Cách nữa.Phá wall cả bác nữa đấy Hiếu 3d

Theo cauchy thần thánh:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\dfrac{\left(a+b\right)\left(a+b\right)}{ab\left(a+b\right)}\ge\dfrac{4ab}{ab\left(a+b\right)}\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{ab\left(a+b\right)}\ge\dfrac{4ab}{ab\left(a+b\right)}\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow\sqrt{\left(a+b\right)^2}\ge\sqrt{4ab}\)

\(\Rightarrow a+b\ge2\sqrt{ab}\)

Ta có điều phải chứng minh

Vâng theo em còn 1 cách nữa bác Hiếuạ(quẩy tung wall bác kia luôn)

Theo cauchy thần thánh: \(\left(a+b\right)^2\ge4ab\)

Từ đó có thể biến đổi:

\(\sqrt{\left(a+b\right)^2}\ge\sqrt{4ab}\)

\(\Rightarrow a+b\ge2\sqrt{ab}\)

Có điều phải chứng minh

Cách khác:

Giả sử:

\(a+b\ge2\sqrt{ab}\)

\(\Rightarrow\dfrac{a+b}{2}\ge\dfrac{2\sqrt{ab}}{2}\)

\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge\left(\dfrac{2\sqrt{ab}}{2}\right)^2\)

\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge\dfrac{4ab}{4}\)

\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\)(luôn đúng theo cauchy)

vậy \(a+b\ge2\sqrt{ab}\)

a)Theo bất đẳng thức cauchy:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{4}{a+b}.\left(a+b\right)\)

\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)

Dấu "=" xảy ra khi: \(a=b\)

Ta có điều phải chứng minh

b)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge\dfrac{9}{a+b+c}.\left(a+b+c\right)\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge9\)

Dấu "=" xảy ra khi:

\(a=b=c\)

Ta có điều phải chứng minh

\(\left\{{}\begin{matrix}x\left(x+y+z\right)=13\\y\left(x+y+z\right)=7\\z\left(x+y+z\right)=-4\end{matrix}\right.\) \(\Leftrightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=13+7-4\)

\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=16\)

\(\Rightarrow\left(x+y+z\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x+y+z=4\\x+y+z=-4\end{matrix}\right.\)

Với \(x+y+z=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=\dfrac{7}{4}\\z=-1\end{matrix}\right.\)

Với \(x+y+z=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{13}{4}\\y=-\dfrac{7}{4}\\z=1\end{matrix}\right.\)

Theo đề bài thì:

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)

\(=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}\)

\(=\dfrac{\left(a+b+b+c+c+a\right)-a-b-c}{c+a+b}\)

\(=\dfrac{a+b+c}{c+a+b}=1\)

Nên: \(\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\)

Mà

\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)

\(P=\left(\dfrac{a}{a}+\dfrac{b}{a}\right)\left(\dfrac{b}{b}+\dfrac{c}{b}\right)\left(\dfrac{c}{c}+\dfrac{a}{c}\right)\)

\(P=\left(\dfrac{a+b}{a}\right)\left(\dfrac{b+c}{b}\right)\left(\dfrac{c+a}{c}\right)\)

\(P=\left(\dfrac{b+c-a+c+a-b}{a}\right)\left(\dfrac{c+a-b+a+b-c}{b}\right)\left(\dfrac{a+b-c+b+c-a}{c}\right)\)

\(P=\dfrac{2c}{a}.\dfrac{2a}{b}.\dfrac{2b}{c}=\dfrac{8ab}{abc}=8\)

Vậy \(P=8\)

Giả sử:

\(a+b\ge2\sqrt{ab}\)

Vì cả 2 vế không âm nên:

\(\left(a+b\right)^2\ge\left(2\sqrt{ab}\right)^2\)

\(\Rightarrow a^2+2ab+b^2\ge4ab\)

\(\Rightarrow a^2+2ab+b^2-2ab=4ab-2ab\)

\(\Rightarrow a^2+b^2\ge2ab\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

Dấu"=" xảy ra khi:

\(\left(a-b\right)^2=0\Rightarrow a=b\)

Điều giả sử luôn đúng nên:

\(a+b\ge2\sqrt{ab}\)