Trước tiên ta chứng minh : \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\); ở đó, a,b tùy ý. Thật vậy:

\(2\left(a^2+b^2\right)S\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\)

Ta có: \(x\left(x-1\right)+\frac{1}{4}+y\left(y-1\right)+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\ge\frac{1}{2}\text{[}\left(x-\frac{1}{2}\right)+\left(y-\frac{1}{2}\right)\text{]}^2\)\(=\frac{1}{2}\left(x+y-1\right)^2\ge\frac{1}{2}\left(6-1\right)^2=\frac{25}{2}\Rightarrow x\left(x-1\right)+y\left(y-1\right)\ge\frac{25}{2}-\frac{1}{2}=12\)

Khi x=y=3 thì => đpcm

Hơi khác cách của mình nhưng đúng r.

Ta có: 

\(2x=3y=6z\)

\(\Rightarrow\)\(\frac{2x}{6}=\frac{3y}{6}=\frac{6z}{6}\)

\(\Rightarrow\)\(\frac{x}{3}=\frac{y}{2}=\frac{2z}{2}=\frac{x+y-2z}{3+2-2}=\frac{27}{3}=9\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=9\\\frac{y}{3}=9\\\frac{2z}{2}=9\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=27\\y=18\\z=9\end{cases}}\)

xin cho tui sửa lại tí @@

Ta có: \(2x=3y=6z\)

\(=>\frac{2x}{6}=\frac{3y}{6}=\frac{6z}{6}\)

\(=>\frac{x}{3}=\frac{y}{2}=\frac{2z}{2}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{x+y-2z}{3+2-2}=\frac{27}{3}=9\)

\(=>\hept{\begin{cases}\frac{x}{3}=9=>x=9\cdot3=27\\\frac{y}{2}=9=>y=9\cdot2=18\\\frac{2z}{2}=9=>z=9\cdot2:2=9\end{cases}}\)

Vậy: x = 27

        y = 18

        z = 9

Ta có: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2ab+2bc+2ac}\)

Mặt khác : \(a^2+b^2+c^2\ge ab+bc+ac\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)\(\Rightarrow\frac{\left(a+b+c\right)^2}{2ab+2bc+2ac}\ge\frac{3}{2}\)

Dự đoán \(MinL=\frac{3}{2}\)khi a = b = c

Ta cần chứng minh \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}\Leftrightarrow\left(\frac{a}{a+b}-\frac{1}{2}\right)+\left(\frac{b}{b+c}-\frac{1}{2}\right)+\left(\frac{c}{c+a}-\frac{1}{2}\right)\ge0\)\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}+\frac{b-c}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\Leftrightarrow\frac{a-b}{2\left(a+b\right)}-\frac{\left(a-b\right)+\left(c-a\right)}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}-\frac{a-b}{2\left(b+c\right)}-\frac{c-a}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{a-b}{2}\left(\frac{1}{a+b}-\frac{1}{b+c}\right)-\frac{c-a}{2}\left(\frac{1}{b+c}-\frac{1}{c+a}\right)\ge0\)\(\Leftrightarrow\frac{a-b}{2}.\frac{c-a}{\left(a+b\right)\left(b+c\right)}-\frac{c-a}{2}.\frac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{\left(a-b\right)\left(c-a\right)\left(c+a\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}-\frac{\left(a-b\right)\left(c-a\right)\left(a+b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\Leftrightarrow\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(đúng do \(a\ge b\ge c>0\))

Đẳng thức xảy ra khi a = b = c

Find the mistake and correct it:

- The new student’s progress advanced forward (A) with such (B) speed that all (C) his teachers were amazed (D).

Sai A nhé:)

=>- The new student’s progress advanced (A) with such (B) speed that all (C) his teachers were amazed (D).

Sửa đề bài ( thêm ) . Tìm tất cả các hàm \(f:ℝ\rightarrowℝ\)

Áp dụng bđt thức svacxo: \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}\ge\frac{\left(x_1+x_2\right)^2}{y_1+y_2}\)  (1)

CM bđt đúng: Áp dụng bđt bunhiacopxki, ta có: (với y1; y2 > = 0)

\(\left[\left(\frac{x_1}{\sqrt{y_1}}\right)^2+\left(\frac{x_2}{\sqrt{y_2}}\right)^2\right]\left[\left(\sqrt{y_1}\right)^2+\left(\sqrt{y_2}\right)^2\right]\ge\left(\frac{x_1}{\sqrt{y_1}}.\sqrt{y_1}+\frac{x_2}{\sqrt{y_2}}.\sqrt{y_2}\right)^2\)

\(\ge\left(x_1+x_2\right)^2\) => \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}\ge\frac{\left(x_1+x_2\right)^2}{y_1+y_2}\) (đpcm)

Ta có: \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\) => \(\sqrt{a^2+b^2}\ge\sqrt{\frac{\left(a+b\right)^2}{2}}=\frac{a+b}{\sqrt{2}}\)(Vì a,b > = 0) (1)

CMTT: \(\sqrt{b^2+c^2}\ge\frac{b+c}{\sqrt{2}}\) (2)

\(\sqrt{c^2+a^2}\ge\frac{a+c}{\sqrt{2}}\) (3)

Từ (1) ; (2) và (3) ta có:  \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\frac{a+b}{\sqrt{2}}+\frac{b+c}{\sqrt{2}}+\frac{a+c}{\sqrt{2}}\)

\(S\ge\frac{a+b+b+c+c+a}{\sqrt{2}}=\frac{2\left(a+b+c\right)}{\sqrt{2}}=3\sqrt{2}=\sqrt{18}\)(Đpcm)

Ta chứng minh BĐT Minkowski: \(\sqrt{m^2+n^2}+\sqrt{p^2+q^2}\ge\sqrt{\left(m+p\right)^2+\left(n+q\right)^2}\)(*)

Thật vậy: (*)\(\Leftrightarrow\left(m^2+n^2\right)+\left(p^2+q^2\right)+2\sqrt{\left(m^2+n^2\right)\left(p^2+q^2\right)}\ge m^2+p^2+2mp+n^2+q^2+2nq\)\(\Leftrightarrow\left(m^2+n^2\right)\left(p^2+q^2\right)\ge\left(mp+nq\right)^2\)(đúng theo BĐT Cauchy-Schwarz)

Áp dụng, ta được: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\sqrt{\left(a+b\right)^2+\left(b+c\right)^2}+\sqrt{c^2+a^2}\)\(\ge\sqrt{\left(a+b+c\right)^2+\left(b+c+a\right)^2}=\sqrt{3^2+3^2}=\sqrt{18}\)

Đẳng thức xảy ra khi a = b = c = 1