Inequality 2

If a,b,c and d are real positive such that :
a + b + c + d = 1
Show that :
√4a+1 + √4b+1 + √4c+1 + √4d+1 < 6

3 Answers

By power mean inequality

(4a+1) + ...+ (4d+1)4 â‰¥ (âˆš4a+1 + ...+âˆš4d+1 4)²

giving 6> 4âˆš2â‰¥âˆš4a+1 + ...+âˆš4d+1

Also:

Make the transformation a\rightarrow \frac{\tan^2 a}{4}

So the Inequality remains to prove \sum{\sec a}<6

By Tchebycheff Inequality, \left( \sum{ \sec a} \right)^2<4\left( \sum{ \sec ^2 a} \right)=32

The inequality follows

Thanks to pidhopit and devil for subscribing in the topic

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω