Ta cần chứng minh \(\dfrac{a^3}{1+b^2}+\dfrac{b^3}{1+a^2}\ge1\)

\(\Leftrightarrow\dfrac{a^3}{ab+b^2}+\dfrac{b^3}{ab+a^2}\ge1\) \(\Leftrightarrow\dfrac{a^3}{b\cdot\left(a+b\right)}+\dfrac{b^3}{a\left(a+b\right)}\ge1\) \(\Leftrightarrow\dfrac{a^4+b^4}{ab\left(a+b\right)}\ge1\Leftrightarrow\dfrac{a^4+b^4}{a+b}\ge1\) 

Áp dụng bđt Cô-si vào 2 số a,b>0 :

 \(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\a^4+b^4\ge2a^2b^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\cdot\left(a^2+b^2\right)\ge\left(a+b\right)^2\\2\cdot\left(a^4+b^4\right)\ge\left(a^2+b^2\right)^2\end{matrix}\right.\) \(\Rightarrow a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(a+b\right)^4}{8}\)

\(\Rightarrow\dfrac{a^4+b^4}{a+b}\ge\dfrac{\left(a+b\right)^3}{8}\ge\dfrac{\left(2\sqrt{ab}\right)^3}{8}=1\) 

Dấu bằng xảy ra \(\Leftrightarrow a=b=1\) Vậy...

Ta có:ab=1⇔a=\(\dfrac{1}{b}\)

Thay a=\(\dfrac{1}{b}\) vào \(\dfrac{a^3}{1+b^2}+\dfrac{b^3}{1+a^2}\) có

\(\dfrac{\left(\dfrac{1}{b}\right)^3}{1+b^2}+\dfrac{b^3}{1+\left(\dfrac{1}{b}\right)^2}\)=\(\dfrac{\left(\dfrac{1}{b}\right)^3}{1+b^2}+\dfrac{b^3}{\dfrac{b^2+1}{b^2}}\)=\(\dfrac{\left(\dfrac{1}{b}\right)^3}{1+b^2}+\dfrac{b^5}{1+b^2}\)=\(\dfrac{\left(\dfrac{1}{b}\right)^3+b^5}{1+b^2}\)=\(\dfrac{\dfrac{1+b^8}{b^3}}{1+b^2}\)

Mà b là số thực dương nên \(\dfrac{\dfrac{1+b^8}{b^3}}{1+b^2}\)≥1

vậy \(\dfrac{a^3}{1+b^2}+\dfrac{b^3}{1+a^2}\)≥1

Theo BĐT Cô - si có : \(a+b\ge2\sqrt{ab}=2\Rightarrow\left(a+b\right)^3\ge8\)

Áp dụng BĐT Svac-xơ ta có :

\(\dfrac{a^3}{1+b^2}+\dfrac{b^3}{1+a^2}=\dfrac{a^4}{a+ab^2}+\dfrac{b^4}{b+a^2b}\)

\(\ge\dfrac{\left(a^2+b^2\right)^2}{a+b+ab.\left(a+b\right)}\ge\dfrac{\left[\dfrac{\left(a+b\right)^2}{2}\right]^2}{a+b+a+b}\) \(=\dfrac{\dfrac{\left(a+b\right)^4}{4}}{2.\left(a+b\right)}=\dfrac{\left(a+b\right)^3}{8}\ge\dfrac{8}{8}=1\)

Dấu "=" xảy ra khi \(a=b=1\)

Đặt\(P=\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+a^2+}+\dfrac{1}{2}\left(ab+bc+ca\right)\) 

Bổ đề: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\) \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) (1)

Chứng minh bổ đề: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\sqrt[3]{abc.\dfrac{1}{abc}}=9\left(\forall a,b,c\ge0\right)\) 

Kết hợp điều kiện đề bài ta được: \(a+b+c\ge3\)

Ta có: \(\dfrac{ab^2}{1+b^2}\le\dfrac{ab^2}{2\sqrt{b^2}}=\dfrac{ab}{2}\) ( AM-GM cho 2 số không âm 1 và b^2 )

\(\Rightarrow\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab}{2}\left(1\right)\)

Chứng minh hoàn toàn tương tự: \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2}\left(2\right)\)

\(\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\left(3\right)\)

Cộng (1),(2),(3) vế theo vế thu được: \(P\ge a+b+c=3\)

Dấu "=" xảy ra tại a=b=c=1

Cách giải của

C113

Ta có: \(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{1}{{a + b + c}} \Longrightarrow \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{{a + b + c}} - \dfrac{1}{c}\)

\(\begin{array}{l} \Longrightarrow \left( {a + b} \right)\left( {a + b + c} \right)c = abc - ab\left( {a + b + c} \right)\\ \Longrightarrow \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) = 0 \end{array}\)

............

C112:

a¹⁶ + a⁸b⁸ + b¹⁶ 

= a¹⁶ + 2a⁸b⁸ + b¹⁶ - a⁸b⁸

= (a⁸ + b⁸)² - (a⁴b⁴)²

= (a⁸ + b⁸ - a⁴b⁴)(a⁸ + b⁸ + a⁴b⁴)

C112 : \(a^{16}+a^8b^8+b^{16}\)

\(=\left(a^8+b^8\right)^2-\left(a^4b^4\right)^2\)

\(=\left(a^8+b^8+a^4b^4\right)\left(a^8+b^8-a^4b^4\right)\)

\(=\left[\left(a^4+b^4\right)^2-\left(a^2b^2\right)^2\right].\left(a^8+b^8-a^4b^4\right)\)

\(=\left(a^4+b^4-a^2b^2\right)\left(a^4+b^4+a^2b^2\right)\left(a^8+b^8-a^4b^4\right)\)

\(=\left[\left(a^2+b^2\right)^2-\left(ab\right)^2\right]\left(a^4+b^4-a^2b^2\right)\left(a^8+b^8-a^4b^4\right)\)

\(=\left(a^2+b^2+ab\right)\left(a^2+b^2-ab\right)\left(a^4+b^4-a^2b^2\right)\left(a^8+b^8-a^4b^4\right)\)

Dear my friend!

As you know that since the outbreak of Covid, we have suffered great consequences and many people have suffered losses to their lives and many other things !! In my country, the influence of Covid-19 is not at all light, when I translated Covid to my city, I was very worried and impatient. The government has told and advised people not to go out, and wear masks ... Then then we started to be more careful when going out, to wear masks, That made me very is annoying because it was not very comfortable and can be said that it messed up my life and everyone in the country, and I had to learn through Zoom, after all that time I found it ineffective because studying on Zoom makes me not very focused and makes the lesson difficult to understand for me, what I tell you is true and if possible I want to advise everyone to keep health because we still have to fight this epidemic is quite long and everyone should follow all the rules that the government, everyone has set out to repel this epidemic.

  I hope I will speak with you soon  !!

Dear my friend!

Surely you also know about the Covid 19 epidemic and its heavy effects, right? To be honest, in Vietnam, we are not affected much by the pandemic, but we are also aware of its danger, so we take precautions by wearing masks and hand washing antiseptic. But recently, in Hai Duong, there were cases of infection in the community, and gradually, more and more cases of infections appeared, and more and more people were F1, F2 (people in contact with Covid 19 infected people.) more. Not only that, the epidemic broke out on the occasion of the Lunar New Year, so we must be more conscious of preventing the spread of epidemic. Students must study online, workers struggle on difficult days, and doctors work day and night to fight the epidemic. Even so, there are still some people who are unconscious and take advantage of it to do bad things. Bad deeds deserve condemnation and good deeds deserve praise. So, let's join hands together to repel translation, so that our life can return to normal as before ! Let's fight off a terrifying plague together !

Hope we will have a chance to talk longer! See you again !

Currently, the Covid-19 epidemic is always a hot topic of concern for the whole world, because it is not only dangerous to its life, but also infects at a dizzying rate. As long as you come into contact with an infected person without wearing a mask or medical equipment, the possibility of infection is very high. In my country, everyone was united in disease prevention, I was extremely surprised by the warm love that the government gave to the people. What is free food delivery to the people of Truc Bach, what is to prepare sufficient clean rooms for people in isolated areas ... That's what our people are! In Europe, they only treat people with high living ability due to lack of human love, not enough beds, high-cost equipment for treatment, but Vietnam is prepared by the Prime Minister. Without orders from their hearts, they would not have entered the epidemic zone to give hope to the people far away from their homeland. Also because of the Corona pandemic, I and my friends have to stay at home to study online. I feel that studying at home is not as effective as studying in class, so I need to pay attention in each lesson on Zoom. I hope we can conquer the plague thanks to our will, top and bottom as one. Vietnam wins!

C108: Thấy cái này hay hay nên chăm hơn chứ lười quá :v

Đặt \(xy=t\Rightarrow x^2+y^2=4-2t\).

Ta cần chứng minh \(t\left(4-2t\right)\le2\). (*)

Thật vậy \((*)\Leftrightarrow 2(t-2)^2\geq 0\) (luôn đúng).

Đẳng thức xảy ra khi và chỉ khi \(xy=2\) tức x = y =1

C108 :

Áp dụng BĐT Cô - si ta có :

\(xy\left(x^2+y^2\right)=\dfrac{1}{2}\cdot\left[2xy.\left(x^2+y^2\right)\right]\le\dfrac{1}{2}\cdot\left(\dfrac{2xy+x^2+y^2}{2}\right)^2=\dfrac{1}{2}\cdot\dfrac{\left(x+y\right)^4}{4}=\dfrac{1}{2}\cdot\dfrac{2^4}{4}=2\)

Dấu "=" xảy ra khi \(x=y=1\)

C108: Thấy cái này hay hay nên chăm hơn chứ lười quá :v

Đặt \(xy=t\Rightarrow x^2+y^2=4-2t\).

Ta cần chứng minh \(t\left(4-2t\right)\le2\). (*)

Thật vậy \((*)\Leftrightarrow 2(t-2)^2\geq 0\) (luôn đúng).

Đẳng thức xảy ra khi và chỉ khi \(xy=2\) tức x = y =1

Chưa like anh ạ=))

de nhu the ai nhin dc :))))

C96 trùng C94 rồi

Đặt:

x = b + c; y = c + a; z = a + b

=> 2a = y + z - x;

2b = x + z - y;

2c = x + y - z.

Đặt vế trái đề bài là (1),

(1) sẽ trở thành:

½[(y + z - x)/x + 25(x + z - y)/y + 4(x + y - z)/z]

= ½(y/x + 25x/y    +    z/x + 4x/z    +    25z/y + 4y/z)

Áp dụng BĐT CÔSI ta có:

y/x + 25x/y ≥ 2\(\sqrt{ }\)(25xy/xy) = 10

z/x + 4x/z ≥ 2\(\sqrt{ }\)(4xz/xz) = 4

25z/y + 4y/z ≥ 2\(\sqrt{ }\)(100yz/yz) = 20

(1) trở thành BĐT:

(1) ≥ ½(10 + 4 + 20 - 30) = 2

Đẳng thức xảy ra khi:

y/x = 25x/y; z/x = 4x/z; 25z/y = 4y/z

=> 25x² = y²; 4x² = z²; 25z² = 4y²

=> y/x = 5; z/x = 2; z/y = 2/5

=> x = 2; y = 10; z = 4

=> b + c = 2; c + a = 10; a + b = 4

=> c = 2 - b; a = 4 - b

=> (2 - b) + (4 - b) = 10 => b = -2 < 0 (không thỏa mãn)

Vậy đẳng thức không xảy ra và (1) > 2

c96 ad xem lại thử xem hình như sai đề

[Toán.C93_17.2.2021] rất hay và khó! Đó là câu em gửi anh trên Facebook hồi sáng. Và em cũng là người đầu công khai đưa ra lời giải bài này.

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