Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)

Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)

Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)

\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)

Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*

\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{​​}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)

\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)

Đẳng thức xảy ra khi a = b = c

P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:

1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)

\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)

bài 2 xem có ghi nhầm ko

3a biến đổi tí là xong

b tuong tự bài 1 

Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)

Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3

\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)

\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)

\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)

ta sẽ giết ngươi kí tên dép đờ kiu lờ

1.

C/m bổ đề: \(a^3-b^3\ge\frac{1}{4}\left(a^3-b^3\right)\) với \(\forall a,b\in R,a\ge b\)

\(\Leftrightarrow4a^3-4b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\ge0\)

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

\(\Leftrightarrow3\left(a^2-b^2\right)\left(a+b\right)\ge0\)

\(\Leftrightarrow3\left(a+b\right)^2\left(a-b\right)\ge0\)(đúng)

Theo bài ra: \(a^3-b^3\ge3a-3b-4\)

\(\Leftrightarrow\) Cần c/m: \(\left(a-b\right)^3\ge12a-12b-16\)(1)

Thật vậy:

\(\left(1\right)\)\(\Leftrightarrow\left(a-b\right)^3-12\left(a-b\right)+16\ge0\)

\(\Leftrightarrow\left[\left(a-b\right)^3-8\right]-12\left(a-b-2\right)\ge0\)

\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a-b\right)+4\right]-12\left(a-b-2\right)\ge0\)

\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a+b\right)-8\right]\ge0\)

\(\Leftrightarrow\left(a-b-2\right)^2\left(a-b+4\right)\ge0\) (đúng với mọi a,b thỏa mãn \(a,b\in R,a\ge b\))

2.

\(BĐT\Leftrightarrow\frac{1}{\frac{a+b}{ab}}+\frac{1}{\frac{c+d}{cd}}\le\frac{1}{\frac{a+b+c+d}{\left(a+c\right)\left(b+d\right)}}\)

\(\Leftrightarrow\frac{ab}{a+b}+\frac{cd}{c+d}\le\frac{\left(a+c\right)\left(b+d\right)}{a+b+c+d}\)

\(\Leftrightarrow\frac{ab\left(c+d\right)+cd\left(a+b\right)}{\left(a+b\right)\left(c+d\right)}\le\)\(\frac{ab+ad+bc+cd}{a+b+c+d}\)

\(\Leftrightarrow\frac{abc+abd+acd+bcd}{ac+ad+bc+bd}\le\frac{ab+ad+bc+cd}{a+b+c+d}\)

\(\Leftrightarrow\left(ad+ab+bc+cd\right)\left(ac+ad+bc+bd\right)\ge\)\(\left(a+b+c+d\right)\left(abc+abd+acd+bcd\right)\)

\(\Leftrightarrow\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)

\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) (đúng với mọi a,b,c,d>0)

1.

\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)

\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)

Dấu "=" khi \(a=b=c\)

2.

\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)

Dấu "=" khi \(a=b=c=d\)

Thục Trinh, tran nguyen bao quan, Phùng Tuệ Minh, Ribi Nkok Ngok, Lê Nguyễn Ngọc Nhi, Tạ Thị Diễm Quỳnh,

Nguyễn Huy Thắng, ?Amanda?, saint suppapong udomkaewkanjana

Help me!

Bài thứ hai đó áp dụng bđt cauchy showas là ra rồi sử dụng tch bắc cầu tệ.

ta có:

\(\frac{3}{a+b}+\frac{2}{c+d}+\frac{a+b}{\left(a+c\right)\left(b+d\right)}\ge\frac{3}{a+b}+\frac{2}{c+d}+\frac{4\left(a+b\right)}{\left(a+b+c+d\right)^2}\)

xét hiệu:

\(\frac{3}{a+b}+\frac{2}{c+d}+\frac{4\left(a+b\right)}{\left(a+b+c+d\right)^2}-\frac{12}{a+b+c+d}\)

\(=\frac{3}{a+b}+\frac{2}{c+d}-\frac{8\left(a+b\right)+12\left(c+d\right)}{\left(a+b+c+d\right)^2}\)

đặt a+b=x;c+d=y

\(\Rightarrow\frac{3}{a+b}+\frac{2}{c+d}-\frac{8\left(a+b\right)+12\left(c+d\right)}{\left(a+b+c+d\right)^2}=\frac{3}{x}+\frac{2}{y}-\frac{8x+12y}{\left(x+y\right)^2}\ge\frac{3}{x}+\frac{2}{y}-\frac{8x+12y}{4xy}=\frac{3}{x}+\frac{2}{y}-\frac{2}{y}-\frac{3}{x}=0\)

\(\Rightarrow\frac{3}{a+b}+\frac{2}{c+d}+\frac{4\left(a+b\right)}{\left(a+b+c+d\right)^2}\ge\frac{12}{a+b+c+d}\)

\(\Rightarrow\frac{3}{a+b}+\frac{2}{c+d}+\frac{a+b}{\left(a+c\right)\left(b+d\right)}\ge\frac{12}{a+b+c+d}\)

=>đpcm

dấu "=" xảy ra khi a=b=c=d

1) Áp dụng BĐT AM-GM: \(VT\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9=VP\)

Đẳng thức xảy ra khi $a=b=c.$

2) Từ (1) suy ra \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{3^2}{a+b+c}+\frac{1^2}{d}\ge\frac{\left(3+1\right)^2}{a+b+c+d}=VP\)

Đẳng thức..

3) Ta có \(\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\) với $a,b,c>0.$

Cho $c=1$ ta nhận được bất đẳng thức cần chứng minh.

4) Đặt \(a=x^2,b=y^2,S=x+y,P=xy\left(S^2\ge4P\right)\) thì cần chứng minh $$(x+y)^8 \geqq 64x^2 y^2 (x^2+y^2)^2$$

Hay là \(S^8\ge64P^2\left(S^2-2P\right)^2\)

Tương đương với $$(-4 P + S^2)^2 ( 8 P S^2 + S^4-16 P^2 ) \geqq 0$$

Đây là điều hiển nhiên.

5) \(3a^3+\frac{7}{2}b^3+\frac{7}{2}b^3\ge3\sqrt[3]{3a^3.\left(\frac{7}{2}b^3\right)^2}=3\sqrt[3]{\frac{147}{4}}ab^2>9ab^2=VP\)

6) \(VT=\sqrt[4]{\left(\sqrt{a}+\sqrt{b}\right)^8}\ge\sqrt[4]{64ab\left(a+b\right)^2}=2\sqrt{2\left(a+b\right)\sqrt{ab}}=VP\)

Có thế thôi mà nhỉ:v

Xét M= \(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{a+d}+\frac{d}{a+b}\)

=\(\frac{a\left(a+d\right)+c\left(b+c\right)}{\left(a+d\right)\left(b+c\right)}+\frac{b\left(a+b\right)+d\left(c+d\right)}{\left(a+b\right)\left(c+d\right)}\)

Với x,y>0 có: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

<=>\(\frac{x+y}{xy}\ge\frac{4}{x+y}\)

<=>\(\frac{1}{xy}\ge\frac{4}{\left(x+y\right)^2}\)(1) .Dấu "=" xảy ra <=>x=y>0

Áp dụng bđt (1) có:

\(\frac{a\left(a+d\right)+c\left(b+c\right)}{\left(a+d\right)\left(b+c\right)}\ge\frac{4\left(a^2+ad+bc+c^2\right)}{\left(a+b+c+d\right)^2}\)

\(\frac{b\left(a+b\right)+d\left(c+d\right)}{\left(c+d\right)\left(a+b\right)}\ge\frac{4\left(ab+b^2+dc+d^2\right)}{\left(a+b+c+d\right)^2}\)

Cộng vế với vế có: \(M\ge\frac{4\left(a^2+ad+bc+c^2+ab+b^2+dc+d^2\right)}{\left(a+b+c+d\right)^2}\)

Có \(2\left(a^2+b^2+c^2+d^2+ad+bc+ab+cd\right)-\left(a+b+c+d\right)^2\)

=\(a^2+b^2+c^2+d^2-2ac-2db=\left(a-c\right)^2+\left(b-d\right)^2\ge0\)

=>\(2\left(a^2+b^2+c^2+d^2+ad+bc+ab+cd\right)\ge\left(a+b+c+d\right)^2\)

<=>\(\frac{4\left(a^2+b^2+c^2+d^2+ab+bc+cd+ad\right)}{\left(a+b+c+d\right)^2}\ge2\)

hay \(M\ge2\)

Dấu "=" xảy ra <=> a=b=c=d>0

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