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Ta có \(a+b^2\le\dfrac{a^2+1}{2}+b^2=\dfrac{a^2+2b^2+1}{2}\)
\(\Rightarrow\dfrac{2a^2}{a+b^2}\ge\dfrac{4a^2}{a^2+2b^2+1}=\dfrac{4a^4}{a^4+2b^2a^2+a^2}\). Lập 2 BĐT tương tự rồi áp dụng bất đẳng thức BCS, ta có:
\(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge\dfrac{\left(2a^2+2b^2+2c^2\right)^2}{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2}\) \(=\dfrac{4\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+3}\)\(=\dfrac{4.3^2}{3^2+3}=3\).
Mà \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\) nên ta có đpcm. ĐTXR \(\Leftrightarrow a=b=c=1\)
BĐT cần chứng minh tương đương :
\(\sqrt{\dfrac{a^2+b^2}{2}}-\sqrt{ab}\ge\dfrac{a+b}{2}-\dfrac{2ab}{a+b}\)
\(\Leftrightarrow\dfrac{\dfrac{a^2+b^2}{2}-ab}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a+b\right)^2-4ab}{2\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\dfrac{\left(a-b\right)^2}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{\left(a-b\right)^2}{2\left(a+b\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\right)\ge0\)
ta phải chứng minh;
\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}-\dfrac{1}{2\left(a+b\right)}\ge0\)
\(\Leftrightarrow\)\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\ge\dfrac{1}{2\left(a+b\right)}\)
\(\Leftrightarrow a+b\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\)\(\Leftrightarrow2a+2b-\sqrt{2\left(a^2+b^2\right)}-2\sqrt{ab}\ge0\)
\(\Leftrightarrow\left(a+b-\sqrt{2\left(a^2+b^2\right)}\right)+\left(a+b-2\sqrt{ab}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2-2\left(a^2+b^2\right)}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a+b\right)^2-4ab}{a+b+2\sqrt{ab}}\ge0\)
\(\Leftrightarrow\dfrac{-\left(a-b\right)^2}{a+b+\sqrt{2\left(a^2+b^2\right)}}+\dfrac{\left(a-b\right)^2}{a+b+2\sqrt{ab}}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\right)\ge0\)
ta phải chứng minh
\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}-\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\ge0\)
\(\Leftrightarrow\dfrac{1}{a+b+2\sqrt{ab}}\ge\dfrac{1}{a+b+\sqrt{2\left(a^2+b^2\right)}}\)
\(\Leftrightarrow a+b+2\sqrt{ab}\le a+b+\sqrt{2\left(a^2+b^2\right)}\)
\(\Leftrightarrow2\sqrt{ab}\le\sqrt{2\left(a^2+b^2\right)}\Leftrightarrow\left(a-b\right)^2\ge0\)
Dấu BĐT bị ngược, sửa đề: \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).
Đặt \(b^2=x\left(x>0\right)\Rightarrow a+x=2ax\).
Khi đó ta cần chứng minh:
\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)
Áp dụng BĐT AM-GM:
\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\)
\(\le\dfrac{1}{2a^2x+2ax^2}+\dfrac{1}{2ax^2+2a^2x}\)
\(=\dfrac{2}{2ax\left(a+x\right)}\)
\(=\dfrac{1}{ax\left(a+x\right)}\)
\(=\dfrac{1}{2a^2x^2}\)
Ta thấy: \(a+x\ge2\sqrt{ax}\)
\(\Leftrightarrow2ax\ge2\sqrt{ax}\)
\(\Leftrightarrow ax-\sqrt{ax}\ge0\)
\(\Leftrightarrow\sqrt{ax}\left(\sqrt{ax}-1\right)\ge0\)
\(\Leftrightarrow\sqrt{ax}\ge1\)
\(\Rightarrow ax\ge1\)
Khi đó: \(\dfrac{1}{2a^2x^2}\le\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)
Hay \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).
\(P=\dfrac{4a^2}{4b+2c}+\dfrac{4b^2}{4a+2c}+\dfrac{c^2}{4a+4b}\ge\dfrac{\left(2a+2b+c\right)^2}{8a+8b+4c}\)
\(=\dfrac{\left(2a+2b+c\right)^2}{4\left(2a+2b+c\right)}=\dfrac{1}{4}\left(2a+2b+c\right)\)
Theo BĐT \(AM-GM\) ta có :
\(\dfrac{a}{\sqrt{2b^2+2c^2-a^2}}=\dfrac{\sqrt{3}a^2}{\sqrt{3a^2\left(2b^2+2c^2-a^2\right)}}\ge\dfrac{\sqrt{3}a^2}{\dfrac{2a^2+2b^2+2c^2}{2}}=\dfrac{\sqrt{3}a^2}{a^2+b^2+c^2}\)
Tương tự ta có :
\(\dfrac{b}{\sqrt{2c^2+2a^2-b^2}}\ge\dfrac{\sqrt{3}b^2}{a^2+b^2+c^2}\)
\(\dfrac{c}{\sqrt{2a^2+2b^2-c^2}}\ge\dfrac{\sqrt{3}c^2}{a^2+b^2+c^2}\)
Cộng từng vế BĐT :
\(\Rightarrow VT\ge\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=\sqrt{3}\)
\("="\Leftrightarrow a=b=c\)
b, \(\frac{a^3}{b+2c}+\frac{b^3}{c+2a}+\frac{c^3}{a+2b}\ge1\)
\(\frac{a^4}{ab+2ac}+\frac{b^4}{bc+2ab}+\frac{c^4}{ac+2bc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac+2ac+2ab+2bc}\)( Bunhia dạng phân thức )
mà \(a^2+b^2+c^2\ge ab+bc+ac\)
\(=\frac{\left(ab+bc+ac\right)^2}{3+2\left(ab+ac+bc\right)}=\frac{9}{3+6}=1\)( đpcm )
1.
Điều kiện .
Phương trình tương đương với \\
Với ta có:
.
Suy ra .
Vậy phương trình có nghiệm duy nhất
2.
Đặt
Áp dụng bất đẳng thức Cauchy cho hai số dương và ta có
.
Tương tự , .
Cộng các vế ta có .
Mà nên (ta có đpcm).
