Câu 1:

Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)

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

\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)

\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)

Vì \(\left(a-b\right)^2\ge0\forall a,b\)

\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)

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

Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)

\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)

\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)

\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)

Vì \(\left(a-b\right)^2\ge0\forall a,b\)

\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)

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

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

5 , a³+b³+c³\(\ge\) 3abc

\(\Leftrightarrow\) a³+3a²b+3ab²+b³+c³-3a²b-3ab²-3abc\(\ge\) 0

\(\Leftrightarrow\) (a+b)³+c³-3ab(a+b+c) \(\ge0\)

\(\Leftrightarrow\) (a+b+c)(a²+2ab+b²-ac-bc+c²)-3ab(a+b+c) \(\ge0\)

\(\Leftrightarrow\) (a+b+c)(a²+b²+c²-ab-bc-ca)\(\ge0\) (1)

ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)

(a-b)²+(b-c)²+(c-a)²\(\ge0\)

<=> 2a²+2b²+2c²-2ac-2cb-2ab\(\ge0\)

<=>a²+b²+c²-ab-bc-ac\(\ge\) 0 (3)

Từ (1)(2)(3)=> pt luôn đúng

a)\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

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

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

b,c tương tự

d)Áp dụng bđt AM-GM ta được

\(a^4+a^4+b^4+c^4\ge4\sqrt[4]{a^4a^4b^4c^4}=4a^2bc\)

TT\(\Rightarrow a^4+b^4+b^4+c^4\ge4ab^2c\)

\(a^4+b^4+c^4+c^4\ge4abc^2\)

Cộng vế theo vế ta được \(4\left(a^4+b^4+c^4\right)\ge4\left(a^2bc+ab^2c+abc^2\right)\)

\(\Leftrightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\left(đpcm\right)\)

d)

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

\(\Leftrightarrow a^4+b^4+c^4-a^2bc-ab^2c-abc^2\ge0\)

\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2bc-2ab^2c-2abc^2\ge0\)

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

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

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

Luôn đúng với mọi a , b , c

a.

\(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)

\(\Leftrightarrow2a^4+2b^4\ge a^4+ab^3+a^3b+b^4\)

\(\Leftrightarrow a^4+b^4\ge ab^3+a^3b\)

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

\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)

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

Mà \(a^2+ab+b^2=\left(a^2+2\cdot a\cdot\dfrac{1}{2}b+\dfrac{b^2}{4}\right)+\dfrac{3b^2}{4}=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\)

Suy ra (*) đúng => đpcm

Dấu "=" xảy ra khi a = b

b.

\(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)

\(\Leftrightarrow3a^4+3b^4+3c^4\ge a^4+ab^3+ac^3+a^3b+b^4+bc^3+a^3c+b^3c+c^4\)

\(\Leftrightarrow2a^4+2b^4+2c^4\ge ab^3+a^3b+b^3c+bc^3+ca^3+c^3a\)

\(\Leftrightarrow\left(a^4+b^4\right)+\left(b^4+c^4\right)+\left(c^4+a^4\right)\ge\left(a^3b+ab^3\right)+\left(b^3c+bc^3\right)+\left(c^3a+ca^3\right)\)

Theo câu a. thì điều này đúng

Dấu "=" khi a=b=c

c. Theo bất đẳng thức Cauchy-Schwarz, ta có:

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{\left(a+b+c\right)^2}{b+c+a}=a+b+c\left(dpcm\right)\)

Dấu "=" xảy ra khi \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)

\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)

Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)

\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)

Cộng vế:

\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)

\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

chết đăng nhầm sogy nha

Câu a : \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow\left(a-b\right)^2\ge0\)

a: =>2a^2+2b^2>=a^2+2ab+b^2

=>a^2-2ab+b^2>=0

=>(a-b)^2>=0(luôn đúng)

c: =>3a^2+3b^2+3c^2>=a^2+b^2+c^2+2ab+2bc+2ac

=>2a^2+2b^2+2c^2-2ab-2bc-2ac>=0

=>(a-b)^2+(b-c)^2+(a-c)^2>=0(luôn đúng)

wow

wow, chắc xu học lớp 9

\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(a+b\right)\left(b+c\right)}{64}}=\dfrac{3a}{4}\)

Tương tự:

\(\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge\dfrac{3b}{4}\)

\(\dfrac{c^3}{\left(c+a\right)\left(a+b\right)}+\dfrac{c+a}{8}+\dfrac{a+b}{8}\ge\dfrac{3c}{4}\)

Cộng vế:

\(VT+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu "=" xảy ra khi \(a=b=c\)

a) Theo bđt cauchy ta có:

\(a^3+b^3+b^3\ge3\sqrt[3]{a^3.b^6}=3ab^2\)

\(a^3+a^3+b^3\ge3a^2b\)

công vế theo vế ta có \(3\left(a^3+b^3\right)\ge3ab^2+3a^2b\)

\(\Leftrightarrow a^3+b^3+3\left(a^3+b^3\right)\ge a^3+3a^2b+3ab^2+b^3\)

\(\Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\)

suy ra đpcm

ta luôn có \(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow a^2+b^2+a^2+b^2\ge a^2+2ab+b^2\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

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

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

suy ra đpcm

câu c đề có sai ko vậy