Chủ đề này có 11 trả lời

[Ben Beck]

Bài 1:a,b,c>0.CM;$\sum \frac{a(b+c)}{b^2+bc+c^2}\geq 2$

Bài 2:a,b,c>=0:a+b+c=2.CM $\sum a^2b^2+abc\leq 2$

Bài 3:a,b,c>0 CM:$\sum \sqrt{\frac{a^2}{a^2+b^2}}\leq \frac{3}{\sqrt{2}}$

Bài 4:a,b,c>0 CM: $\sum \frac{a^2+bc}{b^2+bc+c^2}\geq 2$

Bài viết đã được chỉnh sửa nội dung bởi Truong Gia Bao: 22-07-2017 - 08:17

[Hoang Dinh Nhat]

Bài 3:a,b,c>0 CM:$\sum \sqrt{\frac{a^2}{a^2+b^2}}\leq \frac{3}{\sqrt{2}}$

Bài 4:a,b,c>0 CM: $\sum \frac{a^2+bc}{b^2+bc+c^2}\geq 2$

Bài 3: Đặt $(a^2;b^2;c^2)=(x;y;z)$

BĐT cần chứng minh trở thành: $\sum \sqrt{\frac{x}{x+y}}\leq \frac{3}{\sqrt{2}}$

Theo $Cauchy-Schwarz$, ta có:

$(\sum \sqrt{\frac{x}{x+y}})^2\leq ((x+z)+(y+x)+(z+y))(\sum \frac{x}{(x+y)(x+z)})=\frac{4(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}$

Cần chứng minh: $\frac{4(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}\leq \frac{9}{2}$

$\Leftrightarrow 8(x+y+z)(xy+yz+zx)\leq 9(x+y)(y+z)(z+x)$. Đây là một BĐT quen thuộc

Dấu '=' xảy ra khi: $a=b=c=1$

[AnhTran2911]

Problem 4: To prove this inequality , we take four familiar lemmas, Which are $Iran 96$, $AM-GM$, $Chebyshev$ and $Cauchy-Schwarz$ $ inequalities$

Firstly, arcording to $AM-GM$ we have$\frac{a^2+bc}{b^2+bc+c^2}=\frac{(a^2+bc)(ab+bc+ca)}{(b^2+bc+c^2)(ab+bc+ca)}\geq{\frac{4(a^2+bc)(ab+bc+ca)}{(b+c)^2(a+b+c)^2}}$

Hence, we must prove that$\frac{a^2+bc}{(b+c)^2}\geq{\frac{(a+b+c)^2}{2(ab+bc+ca)}}$

Because of the condition of $a,b,c$ , applying $Chebyshev's$ $ inequality$ , we work out that:

$\frac{a^2+bc}{(b+c)^2}\geq{\frac{1}{3}.(a^2+b^2+c^2+ab+bc+ca)(\frac{1}{(b+c)^2}}+\frac{1}{(a+c)^2}+\frac{1}{(a+b)^2})\geq{\frac{(a+b+c)^2}{2(ab+bc+ca)}}$

The ultimate inequality is the result of these identities:

$a^2+b^2+c^2+ab+bc+ca\geq{\frac{2}{3}.(a+b+c)^2}$ ( $Cauchy-Schwarz$)

$\sum{\frac{1}{(b+c)^2}\geq\frac{9}{4(ab+bc+ca)}}$ ($Iran 96$)

We absolutely deplete our solution.

 

Bài viết đã được chỉnh sửa nội dung bởi AnhTran2911: 25-07-2017 - 13:48

[AnhTran2911]

The problem $2$ can make a proof easily by using $p,q,r$ changing variables technique combines $Schur's$ $inequality$. You see that.

Bài viết đã được chỉnh sửa nội dung bởi AnhTran2911: 22-07-2017 - 10:50

[AnhTran2911]

We can use problem $1$ to prove the problem $4$ by using $Vornicu Schur's$ $inequality$

Solution for problem 1: Dividing the first rational element both of top and bottom by $b+c$ , then multiplied by a and using $Cauchy-Schwarz's$ $inequality$ , we check:

$LHS\geq{\frac{(a+b+c)^2}{2(ab+bc+ca)-2abc(\sum{\frac{1}{a+b}})}}\geq{2}$

Simplifying , it follows to show that : $a^2+b^2+c^2+2abc(\sum{\frac{1}{a+b}})\geq{2(ab+bc+ca)}$

But, arcording to $Cauchy-Schwarz's$ $inequality$: $\sum{\frac{1}{a+b}}\geq{\frac{9}{2(a+b+c)}}$

Hence, it is equivalent to prove: $a^2+b^2+c^2+\frac{9abc}{a+b+c}\geq{2(ab+bc+ca)}$ , Which is $Schur's$ $inequality$ in fraction form.

We obtain $Q.E.D$

[Ben Beck]

Problem 4: To prove this inequality , we take four familiar lemmmas, Which are $Iran 96$, $AM-GM$, $Chebyshev$ and $Cauchy-Schwarz$ $ inequalities$

Firstly, arcording to $AM-GM$ we have$\frac{a^2+bc}{b^2+bc+c^2}=\frac{(a^2+bc)(ab+bc+ca)}{(b^2+bc+c^2)(ab+bc+ca)}\geq{\frac{4(a^2+bc)(ab+bc+ca)}{(b+c)^2(a+b+c)^2}}$

Hence, we must prove that$\frac{a^2+bc}{(b+c)^2}\geq{\frac{(a+b+c)^2}{2(ab+bc+ca)}}$

Because of the condition of $a,b,c$ , applying $Chebyshev's$ $ inequality$ , we work out that:

$\frac{a^2+bc}{(b+c)^2}\geq{\frac{1}{3}.(a^2+b^2+c^2+ab+bc+ca)(\frac{1}{(b+c)^2}}+\frac{1}{(a+c)^2}+\frac{1}{(a+b)^2})\geq{\frac{(a+b+c)^2}{2(ab+bc+ca)}}$

The ultimate inequality is the result of these identities:

$a^2+b^2+c^2+ab+bc+ca\geq{\frac{2}{3}.(a+b+c)^2}$ ( $Cauchy-Schwarz$)

$\sum{\frac{1}{(b+c)^2}\geq\frac{9}{4(ab+bc+ca)}}$ ($Iran 96$)

We absolutely deplete our solution.

Chỗ sử dụng AM-GM hình như có ngc dấu a ơi b^2+bc+c^2>=3/4(b+c)^2 mà

[viet9a14124869]

Chỗ sử dụng AM-GM hình như có ngc dấu a ơi b^2+bc+c^2>=3/4(b+c)^2 mà

Đúng rồi mà  , bạn AnhTran2911 dùng AM-GM thế này:

$4(b^2+bc+c^2)(ab+bc+ca)\leq (b^2+2bc+c^2+ab+ac)^2=(b+c)^2(a+b+c)^2$

Chứ không phải dùng $b^2+bc+c^2\geq \frac{3}{4}(b+c)^2$ đâu .

[AnhTran2911]

To prove $\sum{\frac{a^2+bc}{(b+c)^2}}\geq{\frac{(a+b+c)^2}{2(ab+bc+ca)}}$, we have the another solution:

$VornicuSchur's$ $inequality$ gives us $\sum{\frac{1}{(b+c)^2}.(a-b)(a-c)}\geq{0}$

This is equivalent to:   $\sum{\frac{a^2+bc}{(b+c)^2}}\geq{\frac{a}{b+c}}$

So, it is surffice to show that : $\frac{a}{b+c}\geq{\frac{(a+b+c)^2}{2(ab+bc+ca)}}$ ( $C-S$ )

We also have $Q.E.D$

 

[Nguyenhuyen_AG]

Bài 2:a,b,c>=0:a+b+c=2.CM $\sum a^2b^2+abc\leq 2$

 

Bất đẳng thức chặt hơn sau đây vẫn đúng

\[a^2b^2 + b^2c^2+c^2a^2+\frac{11}8abc \leqslant 2.\]

[Nguyenhuyen_AG]

The problem $2$ can make a proof easily by using $p,q,r$ changing variables technique combines $Schur's$ $inequality$. You see that.

 

Bài này dồn biến sẽ là lựa chọn tốt nhất.

[Ben Beck]

cảm ơn mn

[AnhTran2911]

Bài này dồn biến sẽ là lựa chọn tốt nhất.

Vâng ạ