a, \(\dfrac{a^2+2ab+b^2}{4}\ge ab\)

\(\Leftrightarrow\)a^2+2ab+b^2>=4ab

\(\Leftrightarrow\)a^2-2ab+b^2>=0

\(\Leftrightarrow\)(a-b)^2>=0 (luôn đúng)

b,\(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

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

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

Ta có a⁴ + b⁴ - a³ b - ab³ = (a - b)(a³ - b³)

= (a -b)² (a² + ab + b²)

= (a - b)² [\(\frac{3b^2}{4}+\left(a+\frac{b}{2}\right)^2\)]\(\ge0\)

Ta lại có a⁴ + b⁴ \(\ge2a^2b^2\)

Từ đó => 2(a⁴ + b⁴) \(\ge\)ab³ + a³ b + 2 a² b²

\(2\left(a^4+b^4\right)\ge\left(a^2+b^2\right)\cdot\left(a^{ }^2+b^2\right)\ge2ab\cdot\frac{\left(a+b\right)^2}{2}=ab\cdot\left(a+b\right)^2=ab^3+2a^2b^2+a^3b\)

a) Áp dụng bất đẳng thức AM-GM : 

\(\left(a^2+b^2\right)\left(a^2+1\right)\ge2\sqrt{a^2b^2}.2\sqrt{a^2}\ge2ab.2a=4a^2b\)

b) Áp dụng bất đẳng thức :\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x;y>0\)

 \(\frac{1}{a+3b}+\frac{1}{b+2c+a}\ge\frac{4}{a+3b+b+2c+a}=\frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)

Tương tự \(\hept{\begin{cases}\frac{1}{b+3c}+\frac{1}{c+2a+b}\ge\frac{2}{b+2c+a}\\\frac{1}{c+3a}+\frac{1}{a+2b+c}\ge\frac{2}{b+2a+c}\end{cases}}\)

Cộng vế với vế ta được : \(VT+VP\ge2VP\Rightarrow VT\ge VP\)(đpcm)

Mình giải hết cho bạn rùi nek :))

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

\(\Leftrightarrow2a^4+2b^4-a^3b-ab^3-2a^2b^2\ge0\)

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

\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)+\left(a^2-b^2\right)^2\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)+\left(a^2-b^2\right)^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a^2-ab+b^2\right)+\left(a^2-b^2\right)^2\ge0\) \(\forall a;b\)                   \(\left(1\right)\)

Lại có: \(a^2-ab+b^2=\left(a^2-2.a.\frac{b}{2}+\frac{b^2}{4}\right)+\frac{3b^2}{4}\)

                                         \(=\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge0\) \(\forall a;b\)                          \(\left(2\right)\)

Từ (1) và (2) suy ra  \(\left(a-b\right)^2\left(a^2-ab+b^2\right)+\left(a^2-b^2\right)^2\ge0\forall a;b\)

\(\Leftrightarrow2\left(a^4+b^4\right)\ge a^3b+ab^3+2a^2b^2\forall a;b\)

Vậy \(2\left(a^4+b^4\right)\ge a^3b+ab^3+2a^2b^2\) với mọi a;b

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a, Ta có : \(\left(a-b\right)^2\ge0< =>a^2-2ab+b^2\ge0< =>a^2+b^2\ge2ab\)

\(\left(a-c\right)^2\ge0< =>a^2-2ac+c^2\ge0< =>a^2+c^2\ge2ac\)

Cộng theo vế hai bất đẳng thức sau : \(a^2+b^2+a^2+c^2\ge2ac+2ab< =>2a^2+b^2+c^2\ge2a\left(b+c\right)\left(đpcm\right)\)

Dấu = xảy ra khi và chỉ khi \(a=b=c\)

a) 2a² + b² + c² ≥ 2a( b + c )

<=> 2a² + b² + c² ≥ 2ab + 2ac

<=> 2a² + b² + c² - 2ab - 2ac ≥ 0

<=> ( a² - 2ab + b² ) + ( a² - 2ac + c² ) ≥ 0

<=> ( a - b )² + ( a - c )² ≥ 0 ( đúng )

Vậy bđt được chứng minh

Đẳng thức xảy ra <=> a = b = c 

b) a⁴ - a³b - ab³ + b⁴ ≥ 0

<=> a³( a - b ) - b³( a - b ) ≥ 0

<=> ( a - b )( a³ - b³ ) ≥ 0

<=> ( a - b )( a - b )( a² + ab + b² ) ≥ 0

<=> ( a - b )²[ ( a² + ab + 1/4b² ) + 3/4b² ] ≥ 0

<=> ( a - b )²[ ( a + 1/2b )² + 3/4b² ) ≥ 0 ( đúng )

Vậy bđt được chứng minh

Đẳng thức xảy ra <=> a = b 

a) \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)

\(\Leftrightarrow2x^2+2y^2\ge\left(x+y\right)^2\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\left(đúng\right)\)

b) \(x^3+y^3\ge\dfrac{\left(x+y\right)^3}{4}\)

\(\Leftrightarrow4x^3+4y^3\ge\left(x+y\right)^3\Leftrightarrow3x^3+3y^3\ge3x^2y+3xy^2\)

\(\Leftrightarrow3x^2\left(x-y\right)-3y^2\left(x-y\right)\ge0\)

\(\Leftrightarrow3\left(x-y\right)\left(x^2-y^2\right)\ge0\Leftrightarrow3\left(x-y\right)^2\left(x+y\right)\ge0\left(đúng\right)\)

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

\(\Leftrightarrow2x^2+2y^2-x^2-2xy-y^2\ge0\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\)(luôn đúng)

1)Áp dụng Bđt Am-Gm \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\)

2)Áp dụng Am-Gm \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;b^2+c^2\ge2bc;a^2+c^2\ge2ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

=>ĐPcm

3)(a+b+c)²\(\ge\)3(ab+bc+ca)

=>a²+b²+c²+2ab+2bc+2ca\(\ge\)3ab+3bc+3ca

=>a²+b²+c²-ab-bc-ca\(\ge\)0

=>2a²+2b²+2c²-2ab-2bc-2ca\(\ge\)0

=>(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)\(\ge\)0

=>(a-b)²+(b-c)²+(c-a)²\(\ge\)0

4)đề đúng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

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

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