Ta thấy bđt đúng với n=1.

Giả sử bđt đúng với n=k. Ta cần c/m bđt đúng với n=k+1

Thật vậy ta có: \(\left(\frac{a+b}{2}\right)^n\le\frac{a^n+b^n}{2}\Leftrightarrow\left(\frac{a+b}{2}\right)^{k+1}\)\(\le\frac{a^{k+1}+b^{k+1}}{2}\)

                     \(\Leftrightarrow\left(\frac{a+b}{2}\right)^k.\frac{a+b}{2}\le\frac{a^{k+1}+b^{k+1}}{2}\left(1\right)\)

Ta có \(VT\left(1\right)=\left(\frac{a+b}{2}\right)^k.\frac{a+b}{2}\le\frac{a^k+b^k}{2}.\frac{a+b}{2}=\frac{a^{k+1}+a^kb+ab^k+b^{k+1}}{4}\)\(\le\frac{a^{k+1}+b^{k+1}}{2}\)

       \(\Leftrightarrow\frac{a^{k+1}+b^{k+1}}{2}-\frac{a^{k+1}+ab^k+a^kb+b^{k+1}}{4}\ge0\Leftrightarrow\left(a^k-b^k\right)\left(a-b\right)\ge0\left(2\right)\)

Ta chứng minh (2): * Giả sử \(a\ge b\)và giả thiết cho \(a\ge-b\)\(\Leftrightarrow a\ge\left|b\right|\Leftrightarrow a^k\ge\left|b\right|^k\ge b^k\Rightarrow\left(a^k-b^k\right)\left(a-b\right)\ge0\)

                            * Giả sử \(a< b\)và giả sử \(-a< b\)\(\Leftrightarrow\left|a\right|^k< b^k\Leftrightarrow a^k< b^k\Leftrightarrow\left(a^k-b^k\right)\left(a-b\right)\ge0\)

Vậy bđt (2) luôn đúng \(\Rightarrowđpcm\)

Đổi: \(\left(\frac{a+b}{2}\right)^n=\frac{\left(a+b\right)^n}{2^n}=\frac{a^n+b^n}{2^n}\)

Vì: \(a^n+b^n=a^n+b^n\)

\(2^n\ge2\)

=> \(\left(\frac{a+b}{2}\right)^n\le\frac{a^n+b^n}{2}\)

sai thế kia mà 2 k đúng

Áp dụng BĐT Bernoulli ta có:

\(\left(\frac{2x}{x+y}\right)^n=\left(1+\frac{x-y}{x+y}\right)^n\ge1+\frac{n\left(x-y\right)}{x+y}\)

\(\left(\frac{2y}{x+y}\right)^n=\left(1-\frac{x-y}{x+y}\right)^n\ge1-\frac{n\left(x-y\right)}{x+y}\)

Cộng theo vế 2 BĐT trên ta có: 

\(\left(\frac{2x}{x+y}\right)^n+\left(\frac{2y}{x+y}\right)^n\ge2\) Hay \(\frac{a^n+b^n}{2}\ge\left(\frac{a+b}{2}\right)^n\)

Bài này bạn chỉ cần chuyển vế biến đổi thôi là được , mình làm mẫu câu 2) :

\(\frac{a^2}{m}+\frac{b^2}{n}\ge\frac{\left(a+b\right)^2}{m+n}\)

\(\Leftrightarrow\frac{a^2n+b^2m}{mn}-\frac{\left(a+b\right)^2}{m+n}\ge0\)

\(\Leftrightarrow\frac{\left(m+n\right)\left(a^2n+b^2m\right)-\left(a^2+2ab+b^2\right).mn}{mn\left(m+n\right)}\ge0\)

\(\Leftrightarrow\frac{a^2mn+\left(bm\right)^2+\left(an\right)^2+b^2mn-a^2mn-2abmn-b^2mn}{mn\left(m+n\right)}\ge0\)

\(\Leftrightarrow\frac{\left(bm-an\right)^2}{mn\left(m+n\right)}\ge0\) ( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow bm=an\)

Câu 3) áp dụng câu 2) để chứng minh dễ dàng hơn, ghép cặp 2 .

a/ Bạn cứ khai triển biến đổi tương đương thôi (mà làm biếng lắm)

b/ Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\Rightarrow xyz=1\)

\(VT=\frac{x^3yz}{y+z}+\frac{y^3zx}{z+x}+\frac{xyz^3}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

\(VT\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\left(x+y+z\right)\ge\frac{1}{2}.3\sqrt[3]{xyz}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

Áp dụng Buhiacopxki có \(\left(\left(\frac{m}{\sqrt{x}}\right)^2+\left(\frac{n}{\sqrt{y}}\right)^2\right)\left(\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2\right)\ge\left(m+n\right)^2\)

\(\RightarrowĐPCM\)

\(b^2+c^2\le a^2\Leftrightarrow\left(\frac{b}{a}\right)^2+\left(\frac{c}{a}\right)^2\le1\)

Đặt \(\left\{{}\begin{matrix}\left(\frac{b}{a}\right)^2=x\\\left(\frac{c}{a}\right)^2=y\end{matrix}\right.\) \(\Rightarrow x+y\le1\)

\(P=\left(\frac{b}{a}\right)^2+\left(\frac{c}{a}\right)^2+\left(\frac{a}{b}\right)^2+\left(\frac{a}{c}\right)^2=x+y+\frac{1}{x}+\frac{1}{y}\)

\(P=x+\frac{1}{4x}+y+\frac{1}{4y}+\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\ge2\sqrt{\frac{x}{4x}}+2\sqrt{\frac{y}{4y}}+\frac{3}{4}.\frac{4}{\left(x+y\right)}\)

\(P\ge2+\frac{3}{\left(x+y\right)}\ge2+\frac{3}{1}=5\) (đpcm)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\) hay \(\left(\frac{b}{a}\right)^2=\left(\frac{c}{a}\right)^2=\frac{1}{2}\Rightarrow b=c=\frac{a}{\sqrt{2}}\)

b) với mọi a,b,c ϵ R và x,y,z ≥ 0 có :
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(1\right)\)
Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Thật vậy với a,b∈ R và x,y ≥ 0 ta có:
\(\frac{a^2}{x}=\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\left(2\right)\)
⇔\(\frac{a^2y}{xy}+\frac{b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
⇔\(\frac{a^2y+b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
⇔\(\frac{a^2y+b^2x}{xy}.\left(x+y\right)xy\ge\frac{\left(a+b\right)^2}{x+y}.\left(x+y\right)xy\)
⇔\(\left(a^2y+b^2x\right)\left(x+y\right)\ge\left(a+b\right)^2xy\)
⇔\(a^2xy+b^2x^2+a^2y^2+b^2xy\ge a^2xy+2abxy+b^2xy\)
⇔\(b^2x^2+a^2y^2-2abxy\ge0\)
⇔\(\left(bx-ay\right)^2\ge0\)(luôn đúng )
Áp dụng BĐT (2) có:
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}=\frac{\left(a+b+c\right)^2}{x+y+z}\)
Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Ta có:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)} \)
= \(\frac{1}{a^2}.\frac{1}{ab+ac}+\frac{1}{b^2}.\frac{1}{bc+ac}+\frac{1}{c^2}.\frac{1}{ac+bc}\)
=\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\)
Áp dụng BĐT (1) ta có:
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}++\frac{1}{c}\right)^2}{2\left(ab+bc+ac\right)}\)
Mà abc=1⇒\(\left\{{}\begin{matrix}ab=\frac{1}{c}\\bc=\frac{1}{a}\\ac=\frac{1}{b}\end{matrix}\right.\)
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=3\sqrt[3]{\frac{1}{1}}=3\)( BĐT cosi )
⇒\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
⇒\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}.3=\frac{3}{2}\)
Vậy \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Chúc bạn học tốt !!!

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)

\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)

\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)

\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)

Dấu "=" xảy ra khi x=y=z

a/ Đề sai (ko nói đến chuyện nhầm lẫn ở hạng tử thứ 2 lẽ ra là bc), bạn cho \(a=b=c=d=0,1\) là thấy vế trái lớn hơn vế phải

b/ \(\frac{1}{2}xy.2xy\left(x^2+y^2\right)\le\frac{1}{2}.\frac{\left(x+y\right)^2}{4}.\frac{\left(2xy+x^2+y^2\right)^2}{4}=\frac{\left(x+y\right)^6}{32}=\frac{64}{32}=2\)

Dấu "=" xảy ra khi \(x=y=1\)

c/ Bình phương 2 vế:

\(\Leftrightarrow\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+2\left(a^2+b^2+c^2\right)\ge3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}\ge a^2+b^2+c^2\)

Ta có: \(\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}\ge2b^2\) ; \(\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}\ge2c^2\); \(\frac{a^2b^2}{c^2}+\frac{a^2c^2}{b^2}\ge2a^2\)

Cộng vế với vế:

\(2\left(\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}\right)\ge2\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow...\)

Dấu "=" xảy ra khi \(a=b=c\)