Đăng giờ thiêng thật đấy, tắt máy lúc 10:45 thì đăng lúc 10:47 :"))

C31.2:

\(P=\dfrac{1}{2}.2a\left(1-b\right)b\left(1-c\right)c\left(1-2a\right)\le\dfrac{1}{2}\left(\dfrac{2a+1-b+b+1-c+c+1-2a}{6}\right)^6=\dfrac{1}{128}\)

Dấu ''='' xảy ra khi \(a=\dfrac{1}{4};b=c=\dfrac{1}{2}\)

C31.1 Hình như sai đề ạ, thay a=b=1/16 thì sẽ thấy bị sai:(

C28.1: Làm lại câu này là vì cách kia có holder nhìn hơi cấn:v

\(VT=\dfrac{a^4b^2}{a^2b+b}+\dfrac{b^4c^2}{b^2c+c}+\dfrac{c^4a^2}{c^2a+a}\ge\dfrac{\left(a^2b+b^2c+c^2a\right)^2}{a^2b+b^2c+c^2a+a+b+c}\)

Theo bđt Côsi ta có:

\(a^2b+c^2a+\dfrac{1}{bc^2}\ge3\sqrt[3]{\dfrac{a^3c^2b}{c^2b}}=3a\)

\(c^2a+b^2c+\dfrac{1}{ab^2}\ge3c\)

\(a^2b+b^2c+\dfrac{1}{a^2c}\ge3b\)

\(\Rightarrow2\left(a^2b+b^2c+c^2a\right)+\dfrac{1}{bc^2}+\dfrac{1}{ab^2}+\dfrac{1}{ca^2}\ge3\left(a+b+c\right)\)

\(\Leftrightarrow2\left(a^2b+b^2c+c^2\right)+\dfrac{a^2b+b^2c+c^2a}{a^2b^2c^2}\ge3\left(a+b+c\right)\)

\(\Leftrightarrow3\left(a^2b+b^2c+c^2a\right)\ge3\left(a+b+c\right)\Leftrightarrow a^2b+b^2c+c^2a\ge a+b+c\)

=> ..... (Làm như cách kia)

:3

C28.1

Ta có VT=\(\dfrac{a^4b^2}{a^2b+b}+\dfrac{b^4c^2}{b^2c+c}+\dfrac{c^4a^2}{c^2a+a}\ge\dfrac{\left(a^2b+b^2c+c^2a\right)^2}{a^2b+b^2c+c^2a+a+b+c}\)

Vì \(\left(a^2b+b^2c+c^2a\right)^3\ge\left(a+b+c\right)^3\) ( Theo bđt holder)

\(\Leftrightarrow a^2b+b^2c+c^2a\ge a+b+c\)

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

Dấu ''='' xảy ra khi a=b=c

C27.1

Ta có: \(P=a^2+b^2+\dfrac{5}{a+b+1}=\left(a^2+1\right)+\left(b^2+1\right)+\dfrac{5}{a+b+ab+1+1}-2\)

\(\ge\dfrac{\left(a+1\right)^2}{2}+\dfrac{\left(b+1\right)^2}{2}+\dfrac{5}{\left(a+1\right)\left(b+1\right)+1}-2\)

\(\ge2\sqrt{\dfrac{\left(a+1\right)^2\left(b+1\right)^2}{4}}+\dfrac{5}{\left(a+1\right)\left(b+1\right)+1}-2\)

\(=\left(a+1\right)\left(b+1\right)+1+\dfrac{5}{\left(a+1\right)\left(b+1\right)+1}-3\)

\(=\dfrac{\left(a+1\right)\left(b+1\right)+1}{5}+\dfrac{5}{\left(a+1\right)\left(b+1\right)+1}+\dfrac{4\left(a+1\right)\left(b+1\right)+4}{5}-3\)

\(\ge2+\dfrac{4.2\sqrt{a}.2\sqrt{b}+4}{5}-3=2+\dfrac{4.4\sqrt{ab}+4}{5}-3=3\)

Dấu ''='' xảy ra khi và chỉ khi a=b=1

C25: b5: Sử dụng kĩ thuật Côsi ngược dấu:

Ta có: \(\dfrac{1}{2bc^2+1}=1-\dfrac{2bc^2}{2bc^2+1}\ge1-\dfrac{2bc^2}{3\sqrt[3]{b^2c^4}}=1-\dfrac{2\sqrt[3]{bc^2}}{3}\)

Cmtt ta được: \(\dfrac{1}{2ca^2+1}\ge1-\dfrac{2\sqrt[3]{ca^2}}{3};\dfrac{1}{2ab^2+1}\ge1-\dfrac{2\sqrt[3]{ab^2}}{3}\)

\(\Rightarrow VT\ge1-\dfrac{2\sqrt[3]{bc^2}}{3}+1-\dfrac{2\sqrt[3]{ca^2}}{3}+1-\dfrac{2\sqrt[3]{ab^2}}{3}=3-2\left(\dfrac{\sqrt[3]{bc^2}+\sqrt[3]{ca^2}+\sqrt[3]{ab^2}}{3}\right)\)

Ta có: Theo bđt Côsi:

\(\sqrt[3]{bc^2}=\sqrt[3]{b.c.c}\le\dfrac{b+c+c}{3}=\dfrac{b+2c}{3}\)

\(\sqrt[3]{ca^2}=\sqrt[3]{c.a.a}\le\dfrac{c+a+a}{3}=\dfrac{c+2a}{3}\)

\(\sqrt[3]{ab^2}=\sqrt[3]{a.b.b}\le\dfrac{a+b+c}{3}=\dfrac{a+2b}{3}\)

\(\Rightarrow\sqrt[3]{bc^2}+\sqrt[3]{ca^2}+\sqrt[3]{ab^2}\le\dfrac{b+2c+c+2a+a+2b}{3}=a+b+c=3\)

\(\Rightarrow3-2\left(\dfrac{\sqrt[3]{bc^2}+\sqrt[3]{ca^2}+\sqrt[3]{ab^2}}{3}\right)=1\)

\(\Rightarrow VT\ge1\)

Dấu ''='' xảy ra khi a=b=c=1

C5:

\(A=\dfrac{a}{1+b^2c}+\dfrac{b}{1+c^2d}+\dfrac{c}{1+d^2a}+\dfrac{d}{1+a^2b}=\dfrac{a^2}{a+ab^2c}+\dfrac{b^2}{b+bc^2d}+\dfrac{c^2}{c+cd^2a}+\dfrac{d}{d+da^2b}\)

Áp dụng BĐT Cauchy Schwars dạng Engel ta có:

\(A\ge\dfrac{\left(a+b+c+d\right)^2}{a+b+c+d+ab^2c+bc^2d+cd^2a+da^2b}=\dfrac{16}{4+\left(ab+cd\right)\left(bc+ad\right)}\) 

\(\ge\dfrac{16}{4+\left(\dfrac{ab+bc+cd+ad}{4}\right)^2}=\dfrac{16}{4+\left[\dfrac{\left(a+c\right)\left(b+d\right)}{2}\right]^2}\ge\dfrac{16}{4+\left[\dfrac{\left(\dfrac{a+b+c+d}{2}\right)^2}{2}\right]^2}=2\)

Dấu ''='' xảy ra khi và chỉ khi a=b=c=d=1

C8:

1) Làm cách dài:(

Đặt \(t=a+b+c\Rightarrow t^2=a^2+b^2+c^2+2\)

Ta có:A= \(\dfrac{a}{b^2+c^2+2}+\dfrac{b}{c^2+a^2+2}+\dfrac{c}{a^2+b^2+2}=\dfrac{a^2}{at^2-a^3}+\dfrac{b^2}{bt^2-b^3}+\dfrac{c^2}{ct^2-c^3}\)

Áp dụng BĐT Cauchy Schwars dạng Engel, ta được: 

A\(\ge\dfrac{\left(a+b+c\right)^2}{at^2+bt^2+ct^2-a^3-b^3-c^3}=\dfrac{t^2}{t^3-\left(a^3+b^3+c^3\right)}=\dfrac{t^3}{t^4-t\left(a^3+b^3+c^3\right)}=\dfrac{t^3}{t^4-\left(a+b+c\right)\left(a^3+b^3+c^3\right)}\)

Cm: \(\left(a+b+c\right)\left(a^3+b^3+c^3\right)\ge\left(a^2+b^2+c^2\right)^2\)  

Thật vậy: \(a^3+b^3+c^3=\dfrac{a^4}{a}+\dfrac{b^4}{b}+\dfrac{c^4}{c}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a+c+b}\Rightarrow\left(a+b+c\right)\left(a^3+b^3+c^3\right)\ge\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow A\ge\dfrac{t^3}{t^4-\left(a^2+b^2+c^2\right)^2}=\dfrac{t^3}{t^4-\left(t^2-2\right)^2}=\dfrac{t^3}{4t^2-4}\)

Ta cần cm: \(\dfrac{t^3}{4t^2-4}\ge\dfrac{3\sqrt{3}}{8}\)

Thật vậy: \(8t^3-12\sqrt{3}t^2+12\sqrt{3}\ge0\Leftrightarrow4\left(2t+\sqrt{3}\right)\left(t-\sqrt{3}\right)^2\ge0\) (Đúng với mọi \(t\ge\sqrt{3}\))

=> Đpcm.

P/s: Cảm ơn anh Quân giúp đỡ em nha:3