Bài 1:

Áp dụng BĐT AM-GM ta có:

$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$

Cộng theo vế và thu gọn:

$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$

Ta có đpcm.

Bài 2:

$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$

$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$

$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$

Cộng theo vế và rút gọn thu được:

$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$ 

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Bài 3:

Áp dụng BĐT Cauchy-Schwarz:

$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{b+c+c+a+a+b}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

câu 2 này ms làm tức thì nà

đầu tiên t c/m câu phụ \(\left(a-b\right)\left(b-c\right)\left(c-a\right)\le\dfrac{3\sqrt{3}}{2}\)

đặt P =VT ta có \(P\le\left|P\right|=\sqrt{P^2}\)

vậy ta c/m \(P^2\le\dfrac{27}{4}\)

<=> \(\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2\le\dfrac{27}{4}\)

không mất tính tổng wat giả sử \(a\ge b\ge c\) (2)

dễ thấy \(\left(b-c\right)^2\le b^2;\left(c-a\right)^2\le a^2\)

=> c/m :\(a^2b^2\left(a-b\right)^2\le\dfrac{27}{4}\Leftrightarrow4a^2b^2\left(a-b\right)^2\le\dfrac{27}{4}\)

áp dụng AM-GM ta có

\(4a^2b^2\left(a-b\right)^2=\left(2ab\right)\left(2ab\right)\left(a^2-2ab+b^2\right)\le\left[\dfrac{2\left(2ab\right)+\left(a^2-2ab+b^2\right)}{3}\right]^3=\left(\dfrac{a^2+2ab+b^2}{3}\right)^3=\dfrac{\left(a+b\right)^6}{27}\)

mặt khác từ (2) ta có \(a+b\le a+b+c=3\)

=>dpcm

@quay trở lại bài toán áp dụng câu phụ mik vừa ns c2 <=> c/m

\(\left(a^3+b^3+c^3\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{243}{4}\)

nhân 3 cho 2 vế r áp dụng AM-GM

\(\left(a^3+b^3+c^3\right)3\left(a+b\right)\left(a+c\right)\left(c+b\right)\)\(\le\dfrac{\left[a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}{4}=\dfrac{\left(a+b+c\right)^6}{4}=\dfrac{729}{4}\)

=> dpcm

giúp jum t @Neet;@Ace Legona (có cách khác AM-GM thì qá tốt nha!!)

áp dụng BĐT \(\sqrt[3]{\dfrac{a^3+b^3+c^3}{3}}\ge\dfrac{a+b+c}{3}\) và \(\sqrt[3]{\dfrac{a^3+b^3}{2}}\ge\dfrac{a+b}{2}\) (c/m dưới dạng tổng quát)

\(\sqrt[3]{a^2+3}=\sqrt[3]{4}.\sqrt[3]{\dfrac{\dfrac{a^2+1}{2}+1}{2}}\ge\sqrt[3]{4}.\dfrac{\sqrt[3]{\dfrac{a^2+1}{2}}+1}{2}\)

\(\sqrt[3]{b^2+3}=\sqrt[3]{7}.\sqrt[3]{\dfrac{5.\dfrac{b^2+1}{5}+1+1}{7}}\ge\sqrt[3]{7}.\dfrac{5\sqrt[3]{\dfrac{b^2+1}{5}}+1+1}{ }\)

\(\sqrt[3]{c^2+3}=\sqrt[3]{12}.\sqrt[3]{\dfrac{5.\dfrac{c^2+1}{10}+1}{6}}\ge\sqrt[3]{12}.\dfrac{5\sqrt[3]{\dfrac{c^2+1}{10}}+1}{6}\)

đặt P = VT của dpcm,ta đc

\(P\ge\dfrac{1}{\sqrt[3]{2}}\left(\sqrt[3]{\dfrac{a^2+1}{2}}+1\right)+\dfrac{1}{5\sqrt[3]{2}}\left(5\sqrt[3]{\dfrac{b^2+1}{5}}+2\right)+\dfrac{1}{5\sqrt[3]{2}}\left(\sqrt[3]{\dfrac{c^2+1}{10}}+1\right)=\left(\sqrt[3]{\dfrac{a^2+1}{4}+\sqrt[3]{\dfrac{b^2+1}{10}}+\sqrt[3]{\dfrac{c^2+1}{20}}}\right)+\dfrac{8}{5\sqrt[3]{2}}\)

AM-GM bộ 3 số ta được

\(\sqrt[3]{\dfrac{a^2+1}{4}}+\sqrt[3]{\dfrac{b^2+1}{10}}+\sqrt[3]{\dfrac{c^2+1}{20}}\ge3\sqrt[9]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{800}}\)

we c/m \(3\sqrt[9]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{800}}+\dfrac{8}{5\sqrt[3]{2}}\ge\dfrac{23}{5\sqrt[3]{2}}\)

<=>\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge100\)

cắn bút bín đổi ta đc \(\left(a^2+1\right)\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]\ge100\)

áp dụng BĐT cauchy- gì gì đó

\(\left(a^2+1\right)\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]\ge\left[a\left(b+c\right)+\left(bc-1\right)\right]^2=\left(ab+bc+ca-1\right)^2\ge10^2=100\)=> dpcm

dấu = xảy ra <=> a=1,b=2,c=3

p/s:có j sai ns t nha cách làm của t khá rườm rà @@

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)

\(P=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}+\sqrt{\dfrac{zx}{y^2+xy+yz+zx}}+\sqrt{\dfrac{xy}{z^2+xy+yz+zx}}\)

\(P=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{zx}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)

\(P\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right)+\dfrac{1}{2}\left(\dfrac{z}{y+z}+\dfrac{x}{x+y}\right)+\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{3}{2}\)

\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(a=b=c=\sqrt{3}\)