Bài 1: CMR:
\(a,\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(b,\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\) với a+b+c=3
Bài 2: \(a,b,c\in N,a+b+c=2021\)
Tìm GTNN \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Bài 1:
a) Áp dụng bđt Cô - si:
\(\dfrac{a}{b^2}+\dfrac{1}{a}\ge\dfrac{2}{b}\)
Tương tự với 2 phân thức còn lại của vế trái rồi cộng lại, ta có:
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)
=> đpcm
Bài dù a + b + c = 2021 hay 1 số bất kì thì bđt luôn \(\ge\dfrac{3}{2}\). Bạn có thể tham khảo bđt Nesbitt
Bài 2:
\(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\dfrac{2021-\left(b+c\right)}{b+c}+\dfrac{2021-\left(c+a\right)}{c+a}+\dfrac{2021-\left(a+b\right)}{a+b}\)
\(=2021\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)
Áp dụng BĐT Svacxo, ta có
\(P\) ≥ \(\dfrac{9}{2}-3=\dfrac{3}{2}\)
Dấu"=" ⇔ ...
Sau khi đã đi tham khảo 7749 người thì đã cho ra một kết quả:v
Bài 2. \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(P=\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1-3\)
\(P=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)
\(P=\dfrac{(2a+2b+3c)( \dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b})}{2}-3 ≥ \dfrac{9}{2}-3=\dfrac{3}{2}\)
Dấu `"="` xảy ra:
\(\Leftrightarrow \begin{cases} a=b=c\\ a+b+c=2021 \end{cases} \)
\(\Leftrightarrow a=b=c=\dfrac{2021}{3}\)
Vậy \(min \) \(P=\dfrac{3}{2}\) khi \(a=b=c=\dfrac{2021}{3}\)
b) Áp dụng bđt Cô-si ta có:
\(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{1}{9}a\left(2c+a\right)\ge\dfrac{2}{3}\sqrt{\dfrac{a^4}{b}}=\dfrac{2a^2}{3\sqrt{b}}\)
Tương tự rồi cộng từng vế:
\(VT+\dfrac{1}{9}a\left(2c+a\right)+\dfrac{1}{9}b\left(2a+b\right)+\dfrac{1}{9}c\left(2b+c\right)\ge\dfrac{2}{3}\left(\dfrac{a^2}{\sqrt{b}}+\dfrac{b^2}{\sqrt{c}}+\dfrac{c^2}{\sqrt{a}}\right)\ge\dfrac{2}{3}.\dfrac{\left(a+b+c\right)^2}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\dfrac{6}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Mà \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\le3\left(a+b+c\right)=9\)
=> \(\sqrt{a}+\sqrt{b}+\sqrt{c}\le3\)
=> \(VT+\dfrac{1}{9}\left(a^2+b^2+c^2+2ab+2bc+2ca\right)\ge2\)
=> VT + \(\dfrac{1}{9}\left(a+b+c\right)^2\) ≥ 2
=> đpcm
Ta có : \(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{2c+a}{9}+\dfrac{b}{3}\ge3\sqrt[3]{\dfrac{a^3}{b\left(2c+a\right)}.\dfrac{2c+a}{9}.\dfrac{b}{3}}=a\)
Tương tự ta được
\(\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{2a+b}{9}+\dfrac{c}{3}\ge b\);
\(\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{a}{3}+\dfrac{2b+c}{9}\ge c\)
Cộng vế với vế
=> \(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}+\dfrac{2\left(a+b+c\right)}{3}\ge a+b+c\)
=> \(\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\)
