Ta có: \(1=4\left(a+b\right)+\sqrt{ab}\ge4.2\sqrt{ab}+\sqrt{ab}=9\sqrt{ab}\Leftrightarrow\sqrt{ab}\le\dfrac{1}{9}\Leftrightarrow ab\le\dfrac{1}{81}\)

  \(\Rightarrow\dfrac{1}{ab}\ge\dfrac{1}{\dfrac{1}{81}}=81\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{1}{9}\)

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

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

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

a/b = -1

1:-1

2:...

a/b=-1

đúng 100%

Lời giải:

$N=a(b+3c)+5bc=(1-b-c)(b+3c)+5bc$

$=b+3c-b^2-3c^2+bc$

$-N=b^2+3c^2-bc-b-3c$

$-2N=2b^2+6c^2-2bc-2b-6c$

$\geq b^2+5c^2-2b-6c$

$=(b+c-1)^2+(2c-1)^2-2bc-2$

$\geq -2(bc+1)$

Mà $bc\leq \frac{(b+c)^2}{4}\leq \frac{1}{4}$

$\Rightarrow bc+1\leq \frac{5}{4}$

$\Rightarrow -2(bc+1)\geq \frac{-10}{4}$
$\Rightarrow -2N\geq \frac{-10}{4}$

$\Rightarrow N\leq \frac{5}{4}$

Vậy $N_{\max}=\frac{5}{4}$ khi $(a,b,c)=(0,\frac{1}{2}, \frac{1}{2})$

\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)

Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)

\(S=\dfrac{1}{a^3+b^3}+\dfrac{1}{a^2b}+\dfrac{1}{ab^2}\ge\dfrac{1}{a^3+b^3}+\dfrac{4}{a^2b+ab^2}\)

\(S\ge\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{a^2b+ab^2}+\dfrac{1}{a^2b+ab^2}+\dfrac{1}{a^2b+ab^2}\right)+\dfrac{1}{ab\left(a+b\right)}\)

\(S\ge\dfrac{16}{a^3+b^3+3a^2b+3ab^2}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{4}.\left(a+b\right)}=\dfrac{20}{\left(a+b\right)^3}\ge20\)

\(S_{min}=20\) khi \(a=b=\dfrac{1}{2}\)

\(\left(a^3+b^3\right)\left(a+b\right)=ab\left(1-a\right)\left(1-b\right)\)

\(\Leftrightarrow\left(1-a\right)\left(1-b\right)=\left(\dfrac{a^2}{b}+\dfrac{b^2}{a}\right)\left(a+b\right)\ge\left(a+b\right)^2\ge4ab\)

\(\Rightarrow1+ab-4ab\ge a+b\ge2\sqrt{ab}\)

\(\Rightarrow3ab+2\sqrt{ab}-1\le0\)

\(\Leftrightarrow\left(\sqrt{ab}+1\right)\left(3\sqrt{ab}-1\right)\le0\)

\(\Leftrightarrow ab\le\dfrac{1}{9}\)

\(ab+bc+ca=0\Rightarrow2ab+2bc+2ca=0\)

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

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)

Mà \(2ab+2bc+2ca=0\)

\(\Rightarrow a^2+b^2+c^2=0\)

Mà \(\hept{\begin{cases}a^2\ge0\\b^2\ge0\\c^2\ge0\end{cases}}\)

\(\Rightarrow a^2=b^2=c^2=0\)

\(\Rightarrow a=b=c=0\)

\(\Rightarrow P=1^{1945}+0^{1975}+\left(-1\right)^{2016}=2\)

Vậy ...

từ a+b+c = 0 => (a+b+c)²=0 => a²+b²+c²+2ab+2bc+2ac=0

từ ab+bc+ac = 0 => a²+b²+c² =0

=> a=b=c=0

=>P= 3