1) Đặt \(\dfrac{b\sqrt{a-1}+a\sqrt{b-1}}{ab}\) là A

\(\)\(A=\dfrac{\sqrt{a-1}}{a}+\dfrac{\sqrt{b-1}}{b}\)

\(\left(\dfrac{\sqrt{a-1}}{a}\right)^2=\dfrac{a-1}{a^2}=\dfrac{1}{a}-\dfrac{1}{a^2}=\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)\)

\(\Rightarrow\)\(\dfrac{\sqrt{a-1}}{a}=\sqrt{\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)}\)

Tương tự: \(\dfrac{\sqrt{b-1}}{b}=\sqrt{\dfrac{1}{b}\left(\dfrac{1}{b}-1\right)}\)

Áp dụng BĐT Cauchy, ta có:

\(\sqrt{\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)}\le\dfrac{\dfrac{1}{a}+\left(1-\dfrac{1}{a}\right)}{2}=\dfrac{1}{2}\)

Tương tự: \(\sqrt{\dfrac{1}{b}\left(\dfrac{1}{b}-1\right)}\le\dfrac{1}{2}\)

Cộng vế theo vế của 2 BĐT vừa chứng minh, ta được:

\(A\le1\left(đpcm\right)\)

Xét: \(a^2+\dfrac{2}{a^3}=\dfrac{1}{3}a^2+\dfrac{1}{3}a^2+\dfrac{1}{3}a^2+\dfrac{1}{a^3}+\dfrac{1}{a^3}\left(1\right)\)

Áp dụng BĐT Cauchy cho 5 số dương trên, ta có: \(\left(1\right)\ge5\sqrt[5]{\dfrac{1}{3}a^2.\dfrac{1}{3}a^2.\dfrac{1}{3}a^2.\dfrac{1}{a^3}.\dfrac{1}{a^3}}=5\sqrt[5]{\dfrac{1}{27}}=\dfrac{5\sqrt[5]{9}}{3}\left(đpcm\right)\)

Dấu ''='' xảy ra khi và chỉ khi \(\dfrac{1}{3}a^2=\dfrac{1}{a^3}\Leftrightarrow a=\sqrt[5]{3}\)

Bài 2:

Áp dụng bất đẳng thức Cauchy-Shwarz dạng Engel có:

\(\dfrac{4}{x}+\dfrac{1}{4y}=\dfrac{4^2}{4x}+\dfrac{1}{4y}\ge\dfrac{\left(4+1\right)^2}{4\left(x+y\right)}=\dfrac{25}{5}=5\)

Dấu " = " khi x = y = \(\dfrac{5}{8}\)

Đề sai với a = b = 1.

Vì \(0\le a\le b\le c\le1\) nên:

\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)

Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)

Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)

Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)

Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)

Bài 2:

\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)

Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)

\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:

\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)

Thiết lập các BĐT tương tự:

\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)

Dấu "=" không xảy ra nên ta có ĐPCM

Lưu ý: lần sau đăng từng bài 1 thôi nhé !

1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:

\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)

TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)

\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)

Cộng vế với vế ta được:

\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)

Ai lm dc bai 3 chua

Đặt \(\left(x^3;y^3;z^3\right)=\left(a;b;c\right)\left(x,y,z>0\right)\)

\(\Rightarrow xyz=1\)

Ta cần chứng minh

\(\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\le1\)

Áp dụng AM-GM, ta có: \(x^3+y^3+1=\left(x+y\right)\left(x^2-xy+y^2\right)+xyz\)

\(\ge\left(x+y\right)xy+xyz=xy\left(x+y+z\right)\)

\(\Rightarrow\dfrac{1}{x^3+y^3+1}\le\dfrac{1}{xy\left(x+y+z\right)}\)

Tương tự: \(\dfrac{1}{y^3+z^3+1}\le\dfrac{1}{yz\left(x+y+z\right)}\)

\(\dfrac{1}{z^3+x^3+1}\le\dfrac{1}{zx\left(x+y+z\right)}\)

Cộng vế theo vế, ta được

\(....\le\dfrac{1}{x+y+z}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{1}{x+y+z}.\dfrac{x+y+z}{xyz}=\dfrac{1}{xyz}=1\)

Vậy ta có đpcm

Đẳng thức xảy ra khi a=b=c=1

1 slot :v

\(\sqrt{\dfrac{a}{b+c-ta}}=\dfrac{a\sqrt{t+1}}{\sqrt{\left(at+a\right)\left(b+c-ta\right)}}\ge\dfrac{2a\sqrt{t+1}}{at+a+b+c-ta}=\dfrac{2a\sqrt{t+1}}{a+b+c}\)

Làm tương tự, cộng lại và rút gọn

Mysterious Persontran nguyen bao quanDƯƠNG PHAN KHÁNH DƯƠNG