Nguyễn Việt Lâm anh làm bài này giúp em với ạ

Akai Haruma giúp em bài trên với ạ

Akai Haruma cô giúp em với !!!

\(\sqrt{2\left(b+c\right)^2+bc}\le\sqrt{2\left(b+c\right)^2+\frac{1}{4}\left(b+c\right)^2}=\frac{3}{2}\left(b+c\right)\)

\(\Rightarrow\frac{\left(1-c\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\frac{2}{3}.\frac{\left(1-c\right)^2}{\left(b+c\right)}\)

Tương tự ta có:

\(Q\ge\frac{2}{3}\left(\frac{\left(1-c\right)^2}{b+c}+\frac{\left(1-a\right)^2}{a+c}+\frac{\left(1-b\right)^2}{a+b}\right)\)

\(Q\ge\frac{2}{3}.\frac{\left(1-a+1-b+1-c\right)^2}{2\left(a+b+c\right)}=\frac{\left(3-\left(a+b+c\right)\right)^2}{3\left(a+b+c\right)}=\frac{4}{3}\)

\(Q_{min}=\frac{4}{3}\) khi \(a=b=c=\frac{1}{3}\)

c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)

\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}.\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}-1}{a}\right)\)

\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(1+a\right)-\left(1-a\right)}.\frac{\left(\sqrt{1-a^2}-1\right)}{a}=-1\)

M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu

a,

\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)

\(=0\)

b,

\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)

\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)

\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)

\(=\left(a-b\right)2b=2ab-2b^2\)

Mới nghĩ ra 3 câu:

a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)

\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)

\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)

c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)

\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)

Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)

\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)

\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)

d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)

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

Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm

Mn giúp e vs ạ! Thanks!

Áp dụng BĐT Cô-si cho 3 số dương, ta có :

\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(a+c\right)}\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\)

Cần chứng minh : \(\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\ge\frac{9}{2\left(a+b+c\right)^2}\)

hay \(8\left(a+b+c\right)^6\ge729abc\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

Thật vậy, ta có : \(\left(a+b+c\right)^3\ge\left(3\sqrt[3]{abc}\right)^3=27abc\)

\(8\left(a+b+c\right)^3=\left(2\left(a+b+c\right)\right)^3=\left(a+b+b+c+a+c\right)^3\)

\(\ge\left(3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\right)^3=27\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

Nhân từng vế 2 bất đẳng thức trên, ta được đpcm

Dấu "=" xảy ra khi a = b = c 

Vậy ...

2. Áp dụng BĐT Cô-si cho 3 số không âm, ta có : 

\(B\ge3\sqrt[3]{\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(a^3+c^3+1\right)}}\)

Ta có : \(a^3+b^3+1\ge3\sqrt[3]{a^3b^3}=3ab\Rightarrow\sqrt{a^3+b^3+1}\ge\sqrt{3ab}\)

Tương tự : ....

\(\Rightarrow\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(c^3+a^3+1\right)}\ge\sqrt{27a^2b^2c^2}=\sqrt{27}\)

\(\Rightarrow B\ge3\sqrt[3]{\sqrt{27}}=3\sqrt{3}\)

Vậy GTNN của B là \(3\sqrt{3}\)khi a = b = c = 1

Bài 1 : 

Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm 

\(\Rightarrow\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(c+a\right)}\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\Rightarrow\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(c+a\right)}\ge\frac{3}{\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Xét \(\frac{3}{\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm

\(\hept{\begin{cases}\sqrt[3]{abc}\le\frac{a+b+c}{3}\\\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\frac{2\left(a+b+c\right)}{3}\end{cases}}\)

Nhân từng vế :

\(\Rightarrow\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\frac{2\left(a+b+c\right)^2}{9}\)

\(\Rightarrow\frac{3}{\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\ge\frac{27}{2\left(a+b+c\right)^2}\)

Mà \(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(c+a\right)}\ge\frac{3}{\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\Rightarrow\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(c+a\right)}\ge\frac{27}{2\left(a+b+c\right)^2}\left(đpcm\right)\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2(a+b)}+\frac{a^2}{a^2(b+c)}+\frac{b^2}{b^2(c+a)}+\frac{(\sqrt[3]{abc})^2}{2abc}\)

\(\geq \frac{(c+a+b+\sqrt[3]{abc})^2}{c^2(a+b)+a^2(b+c)+b^2(c+a)+2abc}=\frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$