Theo BĐT Cauchy ta có

 a+b>=2*sqrt(a*b)

4+ab>=2*sqrt(4*ab)

==>(a+b)(4+ab)>=2sqrt(ab).2sqrt(4ab)>=8ab

\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

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

\(\Leftrightarrow2\sqrt{ab}\ge0\) (Luôn đúng vì a ≥0; b≥0)

Dấu ''='' xảy ra khi a=b=0

hơn 1 năm rồi không ai làm :'(

a) Áp dụng bđt Cauchy ta có :

\(a+b\ge2\sqrt{ab}\)(1)

\(b+c\ge2\sqrt{bc}\)(2)

\(c+a\ge2\sqrt{ca}\)(3)

Nhân (1), (2), (3) theo vế

=> \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{a^2b^2c^2}=8\sqrt{\left(abc\right)^2}=8\left|abc\right|=8abc\)

=> đpcm

Dấu "=" xảy ra <=> a=b=c

b) Áp dụng bđt AM-GM ta có :

\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}=2\sqrt{c^2}=2c\)

TT : \(\frac{ca}{b}+\frac{ab}{c}\ge2a\); \(\frac{bc}{a}+\frac{ab}{c}\ge2b\)

Cộng vế với vế

=> \(2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\)

=> \(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c\)( đpcm )

Dấu "=" xảy ra <=> a=b=c

Ta có \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\Leftrightarrow a+b-2\sqrt{ab}\ge0\Leftrightarrow\frac{a+b}{2}\ge\sqrt{ab}.\)

a)\(\dfrac{\left(a+b\right)^2}{4}\ge ab\)\(\Leftrightarrow\dfrac{a^2+ab+b^2}{4}\ge0\)\(\Leftrightarrow\dfrac{\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}}{4}\ge0\left(đpcm\right)\)

Vậy \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

b) Áp dụng Cauchy, ta có:

\(\dfrac{bc}{a}+\dfrac{ca}{b}\ge2\sqrt{\dfrac{bc}{a}.\dfrac{ca}{b}}=2c\)

Tương tự: \(\dfrac{ca}{b}+\dfrac{ab}{c}\ge2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2b\)

Cộng vế theo vế các BĐT vừa chứng minh rồi rút gọn ta được đpcm.

c) ta có \(3a+5b=12\Rightarrow a=\dfrac{12-5b}{3}=4-\dfrac{5b}{3}\)

\(\Rightarrow P=ab=\left(4-\dfrac{5b}{3}\right)b=4b-\dfrac{5b^2}{3}\)

\(\Rightarrow15P=60b-25b^2=36-\left(25b^2-60b+36\right)=36-\left(5b-6\right)^2\)

\(\Rightarrow15P\le36\Rightarrow P\le\dfrac{36}{15}=\dfrac{12}{5}\) Vậy GTLN của \(P=\dfrac{12}{5}\) tại \(a=2;b=\dfrac{6}{5}\)

a. Đề bài sai (thực chất là nó đúng 1 cách hiển nhiên nhưng "dạng" thế này nó sai sai vì ko ai cho kiểu này cả)

Ta có: \(abc=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\ge27\)

\(\Rightarrow a^2+b^2+c^2+5abc\ge a^2+b^2+c^2+5.27>>>>>8\)

b. 

\(4=ab+bc+ca+abc=ab+bc+ca+\sqrt{ab.bc.ca}\le ab+bc+ca+\sqrt{\left(\dfrac{ab+bc+ca}{3}\right)^3}\)

\(\sqrt{\dfrac{ab+bc+ca}{3}}=t\Rightarrow t^3+3t^2-4\ge0\Rightarrow\left(t-1\right)\left(t+2\right)^2\ge0\)

\(\Rightarrow t\ge1\Rightarrow ab+bc+ca\ge3\Rightarrow a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\ge3\)

- TH1: nếu \(a+b+c\ge4\)

Ta có: \(ab+bc+ca=4-abc\le4\)

\(\Rightarrow P=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+5abc\ge4^2-2.4+0=8\)

(Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;2;0\right)\) và các hoán vị)

- TH2: nếu \(3\le a+b+c< 4\)

Đặt \(a+b+c=p\ge3;ab+bc+ca=q;abc=r\)

\(P=p^2-2q+5r=p^2-2q+5\left(4-q\right)=p^2-7q+20\)

Áp dụng BĐT Schur:

\(4=q+r\ge q+\dfrac{p\left(4q-p^2\right)}{9}\Leftrightarrow q\le\dfrac{p^3+36}{4p+9}\)

\(\Rightarrow P\ge p^2-\dfrac{7\left(p^3+36\right)}{4p+9}+20=\dfrac{3\left(4-p\right)\left(p-3\right)\left(p+4\right)}{4p+9}+8\ge8\)

(Dấu "=" xảy ra khi \(a=b=c=1\))

theo BĐT cô - si ta có :

\(\frac{a+b}{2}\ge\sqrt{ab}\) \(\left(a\ge0,b\ge0\right)\)

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

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

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

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

\(\Leftrightarrow\)\(\frac{1}{4}\cdot2\cdot\left(a+b\right)\ge\frac{1}{4}\cdot\left(\sqrt{a}+\sqrt{b}\right)^2\)

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

\(\Leftrightarrow\)\(\sqrt{\frac{a+b}{2}}\ge\frac{\sqrt{a}+\sqrt{b}}{2}\) \(\left(đpcm\right)\)

Biến đổi tương đương đi

BĐT \(\Leftrightarrow\frac{a+b}{2}\ge\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\Leftrightarrow\frac{a-2\sqrt{ab}+b}{4}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{4}\ge0\)(đúng)

Đẳng thức xảy ra khi a = b

P/s: em ko chắc..

\(a+b+\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)

\(\Rightarrow a+b+\dfrac{1}{4}+\dfrac{1}{4}-\sqrt{a}-\sqrt{b}\ge0\)

\(\Rightarrow\left(a-\sqrt{a}+\dfrac{1}{4}\right)+\left(b-\sqrt{b}+\dfrac{1}{4}\right)\ge0\)

\(\Rightarrow\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\left(\sqrt{b}-\dfrac{1}{2}\right)^2\ge0\) (đúng)

\("="\Leftrightarrow a=b=\dfrac{1}{4}\)