a>1; b>1/2; c>1/3 và \(\frac{1}{a} + \frac{2}{2b + 1} + \frac{3}{3c + 2} \ge 2 \quad (1)\)

Đặt x=a

=>x>1

=>x-1>0

Đặt y=2b+1

=>y>2*1/2+1=1+1=2

=>y-2>0

Đặt z=3c+2

=>z>3

=>z-3>0

P=(a-1)(2b-1)(3c-1)

=(x-1)(y-2)(z-3)

(1) =>\(\frac{1}{x} + \frac{2}{y} + \frac{3}{z} \ge 2 \quad (2)\)

=>\(\frac{1}{x} \ge 2 - \frac{2}{y} - \frac{3}{z} = \left(1 - \frac{2}{y}\right) + \left(1 - \frac{3}{z}\right) = \frac{y - 2}{y} + \frac{z - 3}{z}\)

=>\(\frac{1}{x} \ge \frac{y - 2}{y} + \frac{z - 3}{z} \ge 2\sqrt{\frac{(y - 2)(z - 3)}{yz}} \quad (3)\)

Chứng minh tương tự, ta sẽ có:

\(\frac{2}{y} \ge \left(1 - \frac{1}{x}\right) + \left(1 - \frac{3}{z}\right) = \frac{x - 1}{x} + \frac{z - 3}{z} \ge 2\sqrt{\frac{(x - 1)(z - 3)}{xz}} \quad (4)\)

\(\frac{3}{z} \ge \left(1 - \frac{1}{x}\right) + \left(1 - \frac{2}{y}\right) = \frac{x - 1}{x} + \frac{y - 2}{y} \ge 2\sqrt{\frac{(x - 1)(y - 2)}{xy}} \quad (5)\)

Từ (3),(4),(5) suy ra \(\frac{1}{x} \cdot \frac{2}{y} \cdot \frac{3}{z} \ge 2\sqrt{\frac{(y - 2)(z - 3)}{yz}} \cdot 2\sqrt{\frac{(x - 1)(z - 3)}{xz}} \cdot 2\sqrt{\frac{(x - 1)(y - 2)}{xy}}\)

=>\(\frac{6}{xyz}\ge8\cdot\frac{(x - 1)(y - 2)(z - 3)}{xyz}\)

=>\(6\ge8(x-1)(y-2)(z-3)\)

=>\((x-1)(y-2)(z-3)\le\frac{6}{8}=\frac{3}{4}\)

=>P<=3/4

Dấu '=' xảy ra khi \(\begin{cases} \dfrac{x - 1}{x} = \dfrac{y - 2}{y} = \dfrac{z - 3}{z} \\ \dfrac{1}{x} + \dfrac{2}{y} + \dfrac{3}{z} = 2 \end{cases}\)

Đặt \(\dfrac{x - 1}{x}=\dfrac{y - 2}{y}=\dfrac{z - 3}{z}=k\)

=>\(\dfrac{1}{x}=\dfrac{2}{y}=\dfrac{3}{z}=1-k\)

\(\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=2\)

=>1-k+1-k+1-k=2

=>3-3k=2

=>3k=1

=>k=1/3

=>\(\frac{1}{x}=\frac{2}{y}=\frac{3}{z}=1-\frac13=\frac23\)

=>x=3/2 và y=3 và z=9/2

=>a=3/2; b=1; c=5/6

\(\dfrac{1}{1+a}=1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{bc}{\left(1+b\right)\left(1+c\right)}}\)

Tương tự:

\(\dfrac{1}{1+b}\ge2\sqrt{\dfrac{ac}{\left(1+a\right)\left(1+c\right)}}\) ; \(\dfrac{1}{1+c}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+c\right)}}\)

Nhân vế với vế:

\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Rightarrow abc\le\dfrac{1}{8}\)

\(N_{max}=\dfrac{1}{8}\) khi \(a=b=c=\dfrac{1}{2}\)

Ta có \(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}=2\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{\sqrt{ab}}=4\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=4-\dfrac{2}{\sqrt{ab}}\)

Khi đó P = \(\dfrac{1}{\sqrt{ab}}\left(4-\dfrac{2}{\sqrt{ab}}\right)=-2\left(\dfrac{1}{\sqrt{ab}}-1\right)^2+2\le2\)

Dấu "=" khi a = b = 1 

Bạn tham khảo bài làm nhé

Bài 3:

\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)

\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)

-Tham khảo:

\(4M=\dfrac{4}{\left(a+b\right)+\left(a+c\right)}+\dfrac{4}{\left(a+b\right)+\left(b+c\right)}+\dfrac{4}{\left(c+a\right)+\left(b+c\right)}\)

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

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

=> 8M \(\le\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)

\(\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}=2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=8\)

=> \(M\le1\)

Dấu "=" xảy ra <=> a = b = c = 3/4 

\(\dfrac{1}{2a+b+c}=\dfrac{1}{a+a+b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Tương tự:

\(\dfrac{1}{a+2b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)

Cộng vế:

\(M\le\dfrac{1}{16}\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)

\(M_{max}=1\)  khi \(a=b=c=\dfrac{3}{4}\)

\(Q=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)

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

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

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

\(=\dfrac{9}{4}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\)

Áp dụng BĐT B.C.S:

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

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

CMTT:\(\dfrac{1}{b+a^2}\le\dfrac{b+1}{\left(a+b\right)^2}\)

\(\Rightarrow M\le\dfrac{a+b+2}{\left(a+b\right)^2}\)=\(\dfrac{1}{a+b}+\dfrac{2}{\left(a+b\right)^2}\)

\(a+b\ge2\Rightarrow\)\(\dfrac{1}{a+b}\le\dfrac{1}{2};\dfrac{2}{\left(a+b\right)^2}\le\dfrac{1}{2}\)

\(\Rightarrow M\le1\)

Dấu \(''=''\)xảy ra\(\Leftrightarrow a=b=1\)

Lời giải:

$1=a+b+3ab\leq (a+b)+3.\frac{(a+b)^2}{4}$

$\Rightarrow a+b\geq \frac{2}{3}$

$\Rightarrow a^2+b^2\geq \frac{(a+b)^2}{2}=\frac{2}{9}$

\(p=\sqrt{1-a^2}+\sqrt{1-b^2}+\frac{1-(a+b)}{a+b}=\sqrt{1-a^2}+\sqrt{1-b^2}+\frac{1}{a+b}-1\)

\(\leq \sqrt{(1-a^2+1-b^2)(1+1)}+\frac{1}{\frac{2}{3}}-1=\sqrt{2(2-a^2-b^2)}+\frac{1}{2}\)

Mà \(2-a^2-b^2\leq 2-\frac{2}{9}=\frac{16}{9}\)

Do đó:

\(P\leq \sqrt{\frac{32}{9}}+\frac{1}{2}=\frac{3+8\sqrt{2}}{6}\) và đây chính là giá trị max.

SKY WARS:

Đặt $a+b=t$ thì:

$1\leq t+\frac{3}{4}t^2$

$\Leftrightarrow 4\leq 4t+3t^2$

$\Leftrightarrow 3t^2+4t-4\geq 0$

$\Leftrightarrow (3t-2)(t+2)\geq 0$

Vì $t>0$ nên $3t-2\geq 0\Rightarrow t\geq \frac{2}{3}$