a) Áp dụng Cauchy-Schwarz:

\(\left(a+b\right)^2\le\left(1^2+1^2\right)\left(a^2+b^2\right)=2\left(a^2+b^2\right)\)

b) Áp dụng AM-GM:

\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\a^2+c^2\ge2ac\end{matrix}\right.\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2ab+2bc+2ac\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(a^2+b^2+c^2\ge ab+bc+ac\) (cm ở trên r nên khỏi cm lại đi)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow3\left(ab+bc+ac\right)\le\left(a+b+c\right)^2\)

Kết hợp 2 điều trên:\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

a)2(a²+b²) ≥ (a+b)²

⇔ 2a²+2b² ≥ a²+2ab+b²

xét hiệu

⇔ 2a²+2b²-a²-2ab-b² ≥ 0

⇔ a²-2ab+b² ≥ 0

⇔ (a-b)² ≥ 0 (luôn đúng )

=> đpcm

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

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

\(\Leftrightarrow\)(a-b)^2\(\ge\)0 (2)

(2) đúng nên 1 đúng

b )

chứng minh vế 1 3(a^2+b^2+c^2)\(\ge\)(a+b+c)^2

\(\Leftrightarrow\)3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ca\(\ge\)0

\(\Leftrightarrow\)2a^2+2b^2+2c^2-2ab-2ac-2bc\(\ge\)0

\(\Leftrightarrow\)(a-b)^2+(b-c)^2+(c-a)^2\(\ge\)0 luôn đúng

chứng minh vế 2 (a+b+c)^2\(\ge\)3(ab+bc+ca)

\(\Leftrightarrow\)a^2+b^2+c^2-2ab-2ac-2bc\(\ge\)0

cm như trên suy ra đpcm

d/ Đặt \(x=a+b\) , \(y=b+c\) , \(z=c+a\)

thì : \(a=\frac{x+z-y}{2}\) ; \(b=\frac{x+y-z}{2}\) ; \(c=\frac{y+z-x}{2}\)

Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{\frac{x+z-y}{2}}{y}+\frac{\frac{x+y-z}{2}}{z}+\frac{\frac{y+z-x}{2}}{x}\)

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

\(=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)-\frac{3}{2}\ge\frac{1}{2}.6-\frac{3}{2}=\frac{3}{2}\)

b/ \(a^2\left(1+b^2\right)+b^2\left(1+c^2\right)+c^2\left(1+a^2\right)\ge6abc\)

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

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

Vậy bđt ban đầu dc chứng minh.

c/ \(\left(a+b\right)^2\ge4ab\Leftrightarrow\frac{ab}{a+b}\le\frac{a+b}{4}\)

Tương tự : \(\frac{bc}{b+c}\le\frac{b+c}{4}\) ; \(\frac{ac}{a+c}\le\frac{a+c}{4}\)

Cộng theo vế : \(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{a+c}\le\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}\)

c) theo bđt cauchy ta có

\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+1\ge2b\\a^2+1\ge2a\end{matrix}\right.\)

cộng hết lại rút 2 đi \(\Rightarrowđpcm\)

b)theo bđt bunhiacopxki ta có

\(\left(1^2+a^2\right)\left(1^2+b^2\right)\ge\left(1+ab\right)^2\)

\(\Rightarrowđpcm\)

theo bđt cauchy ta có

\(-\left(a^2d^2+b^2c^2\right)\le-2abcd\)

\(\Leftrightarrow a^2c^2-a^2d^2+b^2d^2-b^2c^2\le a^2c^2-2abcd+b^2d^2\)

\(\Leftrightarrow a^2(c^2-d^2)-b^2(c^2-d^2)\le a^2c^2-2abcd+b^2d^2\)

\(\Leftrightarrow(c^2-d^2)\left(a^2-b^2\right)\le(ac-bd)^2\)

\(\Rightarrowđpcm\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

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

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

a/ Từ BĐT ban đầu ta có:

\(2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

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

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\) (đpcm)

b/ Chia 2 vế của BĐT ở câu a cho 9 ta được:

\(\frac{a^2+b^2+c^2}{3}\ge\frac{\left(a+b+c\right)^2}{9}=\left(\frac{a+b+c}{3}\right)^2\) (đpcm)

c/ Cộng 2 vế của BĐT ban đầu với \(2ab+2bc+2ca\) ta được:

\(a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

d/ Áp dụng BĐT ban đầu cho các số \(a^2;b^2;c^2\) ta được:

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

Mặt khác ta cũng có:

\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\ge ab.bc+bc.ca+ab+ca=abc\left(a+b+c\right)\)

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

e/ Chia 2 vế của BĐT ở câu c cho 9 ta được:

\(\frac{\left(a+b+c\right)^2}{9}\ge\frac{ab+bc+ca}{3}\)

Khai căn 2 vế: \(\Rightarrow\frac{a+b+c}{3}\ge\sqrt{\frac{ab+bc+ca}{3}}\)

f/ Áp dụng BĐT ở câu d:

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)=abc\) (do \(a+b+c=1\))

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

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

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

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

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