Căn bậc hai. Căn bậc ba

Ta có a < b + c; b < c + a; c < a + b nên từ a + b + c = 2 suy ra a, b, c < 1.

BĐT cần cm tương đương:

\(\left(a+b+c\right)^2+2abc< 2\left(ab+bc+ca\right)+2\)

\(\Leftrightarrow abc-\left(ab+bc+ca\right)+1< 0\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)< 0\).

Bất đẳng thức trên luôn đúng do a, b, c < 1.

Vậy ta có đpcm.

Đặt vế trái của BĐT là P:

\(P=\sqrt{\left(a+2\right)\left(b+2\right)}+\sqrt{2b.\left(a+1\right)}\)

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

\(P\le\dfrac{1}{2}\left(2a+3b+5\right)=\dfrac{1}{2}.2024=1012\)

Dấu "=" không xảy ra

Áp dụng BĐT BSC:

\(\sqrt{\dfrac{xy}{x+y+2z}}+\sqrt{\dfrac{yz}{y+z+2x}}+\sqrt{\dfrac{zx}{z+x+2y}}\)

\(=\dfrac{2\sqrt{xy}}{\sqrt{\left(1+1+2\right)\left(x+y+2z\right)}}+\dfrac{2\sqrt{yz}}{\sqrt{\left(1+1+2\right)\left(y+z+2x\right)}}+\dfrac{2\sqrt{zx}}{\sqrt{\left(1+1+2\right)\left(z+x+2y\right)}}\)

\(\le\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}+\dfrac{2\sqrt{yz}}{\sqrt{y}+\sqrt{z}+2\sqrt{x}}+\dfrac{2\sqrt{zx}}{\sqrt{z}+\sqrt{x}+2\sqrt{y}}\)

\(\le\dfrac{1}{4}\left(\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{2\sqrt{xy}}{\sqrt{y}+\sqrt{z}}\right)+\dfrac{1}{4}\left(\dfrac{2\sqrt{yz}}{\sqrt{y}+\sqrt{x}}+\dfrac{2\sqrt{yz}}{\sqrt{z}+\sqrt{x}}\right)+\dfrac{1}{4}\left(\dfrac{2\sqrt{zx}}{\sqrt{z}+\sqrt{y}}+\dfrac{2\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}}{\sqrt{y}+\sqrt{x}}+\dfrac{\sqrt{yz}}{\sqrt{z}+\sqrt{x}}+\dfrac{\sqrt{zx}}{\sqrt{z}+\sqrt{y}}+\dfrac{\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{\sqrt{xy}+\sqrt{yz}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}+\sqrt{zx}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}+\sqrt{zx}}{\sqrt{y}+\sqrt{x}}\right)\)

\(=\dfrac{1}{2}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)=\dfrac{1}{2}\)

Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{9}\)

Ta có: \(P=\sqrt{\left(1+x\right)^2}+\sqrt{\left(1-x\right)^2}\)

\(=\left|1+x\right|+\left|1-x\right|\)

\(=1+x+\left|1-x\right|\)

\(=\left[{}\begin{matrix}1+x+1-x\left(x\le1\right)\\1+x+x-1\left(x>1\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}2\\2x\end{matrix}\right.\)

làm rõ \(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)

\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)

\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\) (đúng)

ok thỏa thuận rồi tui làm nửa sau thui nhé :D

Đặt \(a^2=x;b^2=y;c^2=z\) thì ta có:

\(VT=\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{y+z}}+\sqrt{\dfrac{z}{x+z}}\)

Lại có: \(\sqrt{\dfrac{x}{x+y}}=\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)

Tương tự cộng theo vế rồi áp dụng BĐT C-S ta có:

\(VT^2\le2\left(x+y+z\right)\left[\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(y+z\right)\left(y+x\right)}+\dfrac{z}{\left(z+x\right)\left(z+y\right)}\right]\)

\(\Leftrightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

Vì \(VP^2=\dfrac{9}{2}\) nên cần cm \(VT\le \frac{9}{2}\)

\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+yz+xz\right)\)

Can you continue

Cách 1:

Áp dụng bất đẳng thức \(AM-GM\) ta có:

\(Q=x-2\sqrt{2x-1}=x-\sqrt{4\left(2x-1\right)}\ge x-\dfrac{4+2x-1}{2}=-\dfrac{3}{2}\)

Vậy \(Q_{min}=-\dfrac{3}{2}\Leftrightarrow x=\dfrac{5}{2}\)

Cách 2:

\(Q=x-2\sqrt{2x-1}\\ \Leftrightarrow2Q=2x-4\sqrt{2x-1}\\ \Leftrightarrow2Q=\left(2x-1\right)-4\sqrt{2x-1}+1\\ \Leftrightarrow2Q=\sqrt{\left(2x-1\right)^2}-4\sqrt{2x-1}+4-3\\ \Leftrightarrow2Q=\left(\sqrt{2x-1}-2\right)^2-3\\ mà:\left(\sqrt{2x-1}-2\right)^2\ge0\forall x\ge\dfrac{1}{2}\\ \Rightarrow\left(\sqrt{2x-1}-2\right)^2-3\ge-3\forall x\ge\dfrac{1}{2}\\ \Rightarrow2Q_{min}=-3\\ \Leftrightarrow Q_{min}=-\dfrac{3}{2}\Leftrightarrow x=\dfrac{5}{2}\\ VậyQ_{min}=-\dfrac{3}{2}\Leftrightarrow x=\dfrac{5}{2}\)

\(GT\Rightarrow a+b=5\)

\(P=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}=\dfrac{4}{5}\)

Dấu "=" xảy ra khi \(a=b=\dfrac{5}{2}\)

\(\Leftrightarrow a^4+b^4+2a^2b^2-2a^3b-2ab^3\ge0\)

\(\Leftrightarrow\left(a^2+b^2\right)^2-2ab\left(a^2+b^2\right)\ge0\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(a^2+b^2-2ab\right)\ge0\)

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

\(n^2+2002=k^2\Rightarrow k^2-n^2=2002\)

\(\Rightarrow\left(k-n\right)\left(k+n\right)=2002\)

Do \(\left(k-n\right)+\left(k+n\right)=2k\) chẵn nên \(\left(k-n\right)\) và \(\left(k+n\right)\) cùng chẵn

Bạn chỉ cần xét các cặp ước chẵn của 2002

Ta thấy n² chia cho 4 dư 0 hoặc 1 nên n² + 2002 chia cho 4 dư 2 hoặc 3.

Do đó n² + 2002 không thể là số chính phương.